UNIVERSITY  OF  CALIFORNIA 

DEPARTMENT  OF  CIVIL  ENGINEERING 

BERKELEY.  CALIFORNIA 


UNIVERSITY  OF  CALIFORNIA 

DEPARTMENT  OF  CIVIL  ENGINEERING 

BERKELEY.  CALIFORNIA 


WORKS  OF  DR.  D.  B.  STEINMAN 


PUBLISHED   BY 

JOHN  WILEY  &  SONS,  Inc. 

Suspension  Bridges:  Their  Design,  Construction  and  Erection. 
Stresses  in  suspension  bridges. — Types  and  details  of 
construction. — Typical  design  computations. — Erection 
of  suspension  bridges. — Charts  for  expeditious  design. — 
8vo,  204  pages,  59  illustrations,  3  Design  Charts.  Cloth, 
$4.00,  net. 

Concrete  Arches,  Plain  and  Reinforced. 

Authorized  Translation  from  the  Second  Revised  Edition 
by  Prof.  J.  Melan.  Exact  and  approximate  methods  of 
proportioning  concrete  arches. — Graphic  and  analytic 
methods  of  design. — Numerical  examples  and  actual  de- 
signs completely  worked  out. — The  Melan  System  ex- 
plained.— 8vo,  161  pages,  44  figures,  1  chart  for  designing 
reinforced  concrete  sections,  22  full-page  illustrations. 
Cloth,  $2.50,  net. 


Published  by  McGRA  W-HILL  BOOK  CO.,  Inc. 
Theory  of  Arches  and  Suspension  Bridges. 

Authorized  Translation  from  the  Third  Revised  Edition 
by  Prof.  J.  Melan.  Approximate  and  Exact  Theories  for 
the  Stiffened  Suspension  Bridge. — Arched  Ribs. — Framed 
Arches. — Braced  Cable  Suspension  Bridges. — Cantilever 
Arches. — Continuous  Arches. — Combined  Systems. — Ap- 
pendix on  the  Elastic  Theory  applied  to  Masonry  and 
Concrete  Arches. — Bibliography. — 8vo,  303  pages,  119 
figures,  3  folding  plates.  Cloth,  $3.00. 


Published  by  D.  VAN  NOSTRAND  CO. 
Suspension  Bridges  and  Cantilevers. 

Their  Economic  Proportions  and  Limiting  Spans.  Second 
Edition,  Revised.  An  Economic  Study  of  Long  Span 
Bridges. — Empirical  formulae  for  weights  and  costs. — 
Data  and  methods  of  design. — 5  plates,  185  pages.  (Van 
Nostrand's  Science  Series,  No.  127.)  Price,  $0.75. 


Proposed  Hudson  River  Bridge,  Perspective  from  New  York  Shore.    Span  3240  Ft. 
Design  by  G.  Lindenthal,  1921. 

(Frontispiece) 


A  PRACTICAL  TREATISE    -; 

ON 

SUSPENSION    BRIDGES 

THEIR  DESIGN,  CONSTRUCTION  AND  ERECTION 


BY 

D.  P.  STEINMAN,  A.M.,  C.E.,  Ph.D. 

CONSULTING  ENGINEER 

Member  American  Society  of  Civil  Engineers;  Member  American 

Railway  Engineering  Association;  Formerly  Professor 

in  Charge  of  Civil  and  Mechanical  Engineering 

at  the  College  of  the  City  of  New  York 


WITH  APPENDIX: 
DESIGN  CHARTS  FOR  SUSPENSION  BRIDGES 


NEW   YORK 

JOHN  WILEY  &  SONS,  Inc. 

LONDON;  CHAPMAN  &  HALL,  LIMITED 
1922 


Engineering 
Library 


Copyright,  1922 
By  D.  B.  STEINMAN 


. 
V 


PRESS  OF 

BRAUNWORTH   &,  CO. 

BOOK   MANUFACTURERS 

BROOKUYN,    N.    Yt 


PREFACE 


THIS  book  has  been  planned  to  supply  the  needs  of  practi- 
cing engineers  who  may  have  problems  in  estimating,  designing 
or  constructing  suspension  bridges,  and  of  students  who  wish  to 
prepare  themselves  for  work  in  this  field.  The  aim  has  been  to 
produce  an  up-to-date,  practical  handbook  on  the  subject,  dis- 
tinguished by  simplicity  of  treatment  and  convenience  of  appli- 
cation. 

In  the  first  division,  on  Stresses  in  Suspension  Bridges,  the 
formulas  have  been  corrected  to  conform  to  modern  practice, 
and  reduced  to  their  simplest  and  most  convenient  form  for 
direct  application  by  the  designing  engineer.  The  formulas  are 
supplemented  by  curves  for  their  expeditious  solution,  and  by 
alternative  graphical  methods  for  determining  stresses. 

The  second  division,  on  Types  and  Details  of  Construction, 
presents  data  and  illustrations  to  assist  the  designing  engineer 
in  the  selection  of  type  of  suspension  bridge  and  in  the  determina- 
tion of  proportions,  specifications,  and  details  for  the  various 
elements  of  suspension  construction. 

The  third  division,  on  Typical  Design  Computations,  gives 
numerical  examples  of  suspension  designs  of  different  types 
worked  out  by  methods  that  have  proved  most  efficient  in  the 
author's  practice.  The  designing  engineer  will  find  here  the 
formulas  to  be  used  in  each  successive  step  of  the  design,  and 
the  practical  methods  of  applying  them,  with  tabulations,  graphs 
and  short-cuts. 

The  fourth  division,  on  Erection  of  Suspension  Bridges, 
describes  and  illustrates  the  successive  stages  in  the  erection  of 
representative  structures,  from  towers  to  trusses.  The  opera- 
tions of  stringing  wire  cables  are  presented  in  detail,  with  an 
outline  of  the  computations  for  adjustment  and  control. 

iii 


iv  PREFACE 

Methods  of  erecting  eyebar  chains  and  other  types  are  also  de- 
scribed and  illustrated. 

The  Appendix  presents  a  series  of  design  charts,  specially 
devised  for  this  book,  for  the  expeditious  proportioning  of  suspen- 
sion bridges.  These  charts  give  quickly  and  accurately  the 
governing  stresses  throughout  any  span,  saving  the  time  and 
labor  of  applying  the  stress  formulas  otherwise  required. 

The  author  desires  to  express  his  indebtedness  to  his  associate, 
Mr.  Hoi  ton  D.  Robinson,  for  reviewing  the  manuscript  on 
Erection;  and  to  the  Department  of  Plant  and  Structures  of 
New  York  City  for  many  courtesies  extended. 

D.  B.  STEINMAN. 

NEW  YORK  CITY 
August  i,  1922. 


CONTENTS 


CHAPTER  I 
STRESSES  IN  SUSPENSION  BRIDGES 

SECTION  I. — THE  CABLE 

PAGE 

1.  Form  of  the  Cable  for  Any  Loading i 

2.  The  Parabolic  Cable 4 

3.  Unsymmetrical  Spans 6 

4.  The  Catenary 9 

5.  Deformations  of  the  Cable 1 1 

SECTION  II. — UNSTIFFENED  SUSPENSION  BRIDGES 

6.  Introduction 12 

7.  Stresses  in  the  Cables  and  Towers 13 

8.  Deformations  under  Central  Loading 14 

9.  Deformations  under  Unsymmetrical  Loading 15 

10.  Deflections  due  to  Elongation  of  Cable 16 

SECTION  III. — STIFFENED  SUSPENSION  BRIDGES 

11.  Introduction 18 

12.  Assumptions  Used 18 

13.  Fundamental  Relations 21 

14.  Influence  Lines 23 

SECTION  IV. — THREE-HINGED  STIFFENING  TRUSSES 

15.  Analysis 26 

16.  Moments  in  the  Stiffening  Truss 28 

17.  Shears  in  the  Stiffening  Truss 30 

SECTION  V. — TWO-HINGED  STIFFENING  TRUSSES 

18.  Determination  of  the  Horizontal  Tension  H 33 

19.  Values  of  H  for  Special  Cases  of  Loading 38 

20.  Moments  in  the  Stiffening  Truss 41 

v 


VI  CONTENTS 

PAGE 

21.  Shears  in  the  Stiffening  Truss 45 

22.  Temperature  Stresses 48 

23.  Deflections  of  the  Stiffening  Truss 48 

24.  Straight  Backstays 51 

SECTION  VI. — HINGELESS  STIFFENING  TRUSSES 

25.  Fundamental  Relations 53 

26.  Moments  at  the  Towers 56 

27.  The  Horizontal  Tension  H 57 

28.  Values  of  H  for  Special  Cases  of  Loading 57 

29.  Moments  in  the  Stiffening  Truss 59 

30.  Temperature  Stresses 61 

3 1 .  Straight  Backstays 62 

SECTION  VII. — BRACED-CHAIN  SUSPENSION  BRIDGES 

32.  Three-hinged  Type 63 

33.  Two-hinged  Type 65 

34.  Hingeless  Type 67 

CHAPTER  II 
TYPES  AND  DETAILS  OF  CONSTRUCTION 

1.  Introduction — (Classification  Table) 69 

2.  Various  Arrangements  of  Suspension  Spans 72 

3.  Wire  Cables  vs.  Eyebar  Chains , 74 

4.  Methods  of  Vertical  Stiffening 76 

5.  Methods  of  Lateral  Stiffening 77 

6.  Comparison  of  Different  Types  of  Stiffening  Truss 78 

7.  Types  of  Braced-Chain  Bridges 80 

8.  Economic  Proportions  for  Suspension  Bridges 82 

9.  Arrangements  of  Cross-sections 83 

10.  Materials  used  in  Suspension  Bridges 84 

11.  Wire  Ropes  (for  Cables  and  Suspenders) 85 

12.  Parallel  Wire  Cables 89 

13.  Cradling  of  the  Cables 90 

14.  Anchoring  of  the  Cables 91 

15.  Construction  of  Chains 93 

16.  Suspender  Connections — (Cable  Bands  and  Sockets) 96 

17.  Suspension  of  the  Roadway 98 

18.  Construction  of  Stiffening  Trusses 101 

19.  Braced-Chain  Construction 103 

20.  Wind  and  Sway  Bracing no 

21.  Towers 1 13 

22.  Saddles  and  Knuckles 115 

23.  Anchorages 118 


CONTENTS  vii 


CHAPTER  III 

TYPICAL  DESIGN  COMPUTATIONS 
EXAMPLE  i 

Colorations  for  Two-hinged  Suspension  Bridge  with  Straight  Backstays 
(Type  2F.) 

PAGE 

1.  Dimensions 125 

2.  Stresses  in  Cable 125 

3.  Moments  in  Stiffening  Truss 127 

4.  Shears  in.  Stiffening  Truss 130 

5.  Wind  Stresses  in  Bottom  Chords 132 

EXAMPLE  2 

Calculations  for  Two-hinged  Suspension  Bridge  with  Suspended  Side  Spans 

(Type  2S.} 

1.  Dimensions 134 

2.  Stresses  in  Cable 134 

3.  Moments  in  Stiffening  Truss — Main  Span i  6 

4.  Bending  Moments  in  Side  Spans 138 

5.  Shears  in  Stiffening  Truss — Main  Span 139 

6.  Shears  in  Side  Spans 141 

7.  Temperature  Stresses 142 

8.  Wind  Stresses '. 143 

EXAMPLE  3 

Calculations  for  Towers  of  Two-hinged  Suspension  Bridge 
(Type  2S.} 

1.  Dimensions 144 

2.  Movement  of  Top  of  Tower 145 

3.  Forces  Acting  on  Tower 146 

4.  Calculation  of  Stresses 147 

5.  Wind  Stresses. 148 

EXAMPLE  4 

Estimates  of  Cable  and  Wrapping 

1.  Calculation  of  Cable  Wire 149 

2.  Calculation  of  Cable  Diameter 149 

3.  Calculation  of  Wrapping  Wire 149 

4.  Estimate  of  Rope  Strand  Cables 150 

EXAMPLE  5 
Analysis  of  Suspension  Bridge  with  Continuous  Shjjcning  Truss 

1.  Dimensions 150 

2.  Stresses  in  Cables 151 


viii  CONTENTS 

PAGE 

3.  Influence  Line  for  H 152 

4.  Bending  Moments  in  Main  Span 153 

5.  Shears  in  Main  Span 156 

6.  Bending  Moments  in  Side  Spans 158 

7.  Shears  in  Side  Spans 160 

EXAMPLE  6 
Design  of  Anchorage 

1.  Stability  against  Sliding 162 

2.  Stability  against  Tilting 162 

CHAPTER  IV 
ERECTION  OF  SUSPENSION  BRIDGES 

1.  Introduction 163 

2.  Erection  of  the  Towers 163 

3.  Stringing  the  Footbridge  Cables 165 

4.  Erection  of  Footbridges 167 

5.  Parallel  Wire  Cables 169 

6.  Initial  Erection  Adjustments 169 

7.  Spinning  of  the  Cables 172 

8.  Compacting  the  Cables 177 

9.  Placing  Cable  Bands  and  Suspenders 177 

10.  Erection  of  Trusses  and  Floor  System 178 

11.  Final  Erection  Adjustments .- 182 

1 2.  Cable  Wrapping 183 

13.  Erection  of  Wire-rope  Cables 184 

14.  Erection  of  Eyebar-chain  Bridges 186 

15.  Time  Required  for  Erection 189 

APPENDIX 
DESIGN  CHARTS  FOR  SUSPENSION  BRIDGES 

INTRODUCTION 191 

CHART     I. — Bending  Moments  in  Main  Span 193 

CHART    II. — Shears  in  Main  Span 193 

CHART  III. — Moments  and  Shears  in  Side  Spans • 195 

INDEX 199 


A  PRACTICAL  TREATISE 

ON 

SUSPENSION   BRIDGES 

THEIR  DESIGN,  CONSTRUCTION  AND  ERECTION 


CHAPTER  I 
STRESSES  IN  SUSPENSION  BRIDGES 

SECTION  I.— THE  CABLE 

1.  Form  of  the  Cable  for  Any  Loading. — If  vertical  loads 
are  applied  on  a  cable  suspended  between  two  points,  it  will 
assume  a  definite  polygonal  form  determined  by  the  relations 
between  the  loads  (Fig.  id). 

The  end  reactions  (Ti  and  T%)  will  be  inclined  and  will  have 
horizontal  components  H.  Simple  considerations  of  static 
equilibrium  show  that  H  will  be  the  same  for  both  end  reactions, 
and  will  also  equal  the  horizontal  component  of  the  tension  in 
the  cable  at  any  point.  H  is  called  the  horizontal  tension  of  the 
cable 

Let  M'  denote  the  bending  moment  produced  at  any  point 
of  the  span  by  the  vertical  loads  and  reactions,  calculated  as  for 
a  simple  beam.  Since  H,  the  horizontal  component  of  the  end 
reaction,  acts  with  a  lever-arm  yt  the  total  moment  at  any  point 
of  the  cable  will  be 

M  =  M'-H-y (i) 

This  moment  must  be  equal  to  zero  if  the  cable  is  assumed  to  be 
flexible.     Hence, 

M'=H-y, V.   .    .    (2) 


STRESSES   IN   SUSPENSION  BRIDGES 

•  v*'p 


and 


(3) 


Equation  (3)  gives  the  ordinates  to  the  cable  curve  for  any 
loading,  if  the  horizontal  tension  H  is  known.  Since  H  is  con- 
stant, the  curve  is  simply  the  bending  moment  diagram  for  the 
applied  loads,  drawn  to  the  proper  scale.  The  scale  for  con- 


FIG.  i. — The  Cable  as  a  Funicular  Polygon. 

strutting  this  diagram  is  determined  if  the  ordinate  of  any  point 
of  the  curve,  such  as  the  lowest  point,  is  given.  If  /  is  the  sag 
of  the  cable,  or  ordinate  to  the  lowest  point  C,  and  if  Mc  is  the 
simple-beam  bending  moment  at  the  same  point,  then  H  is 
determined  from  Eq.  (2)  by 

„    Mc 

H=7 (4) 

To  obtain  the   cable  curve  graphically,   simply  draw  the 
equilibrium  polygon  for  the  applied  loads,  as  indicated  in  Fig.  i 


THE  CABLE  3 

(a,  b).  The  pole  distance  H  must  be  found  by  trial  or  computa- 
tion so  as  to  make  the  polygon  pass  through  the  three  specified 
points,  Aj  B,  and  C.  The  tension  T  at  any  point  of  the  cable 
is  given  by  the  length  of  the  corresponding  ray  of  the  pole  dia- 
gram. H,  the  horizontal  component  of  all  cable  tensions,  is 
constant.  By  similar  triangles,  the  figure  yields 

*  T=S~=H-SeC4>  ......     (S) 

where  <f>  is  the  inclination  of  the  cable  to  the  horizontal  at  any 
point.  It  should  be  noted  that  the  tensions  T  in  the  successive 
members  of  the  polygon  increase  toward  the  points  of  support 
and  attain  their  maximum  values  in  the  first  and  last  members 
of  the  system. 

If  Vi  is  the  vertical  component  of  the  left  end  reaction,  the 
vertical  shear  at  any  section  x  of  the  span  will  be 


(6) 


This  will  also  be  the  vertical  component  of  the  cable  tension  at 
the  same  point.     By  similar  triangles, 


(7) 


(This  relation  is  also  obtained  by  differentiating  both  members 
of  Eq.  2).     Combining  Eqs.  (6)  and  (7),  we  may  write 


-  (8) 

sr  ~s~- 

If  the  loads  are  continuously  distributed,  the  funicular  poly- 
gon becomes  a  continuous  curve.  If  w  is  the  load  per  horizontal 
linear  unit  at  any  point  having  the  abscissa  x,  Eq.  (8)  becomes 


4  STRESSES'  IN  SUSPENSION  BRIDGES 

from  which  (by  differentiation)  we  obtain  the  following  as  the 
differential  equation  of  the  equilibrium  curve: 

d?y          w  ,    . 


For  any  given  law  of  variation  of  the  continuous  load  wy 
the  integration,  of  Eq.  (10)  will  give  the  equation  of  the  curve 
assumed  by  the  cable. 

2.  The  Parabolic  Cable.  —  For  a  uniform  distributed  load, 
the  bending  moment  diagram  is  a  parabola.  Consequently,  by 
Eq.  (3),  if  a  cable  carries  a  uniform  load  (w  per  horizontal  linear 
unit),  the  resulting  equilibrium  curve  will  be  a  parabola. 

The  maximum  bending  moment  in  a  simple  beam  would  be 

M       Wl2 
Mc=~-. 

Substituting  this  value  in  Eq.  (4),  the  horizontal  tension  is  deter- 
mined : 


To  obtain  the  equation  of  the  curve,  integrate  Eq.   (10). 
With  the  origin  of  coordinates  at  the  crown,  the  integration 

yields 

wx2  ,    N 

y  =  ^H  ........  .  '.<"> 

Substituting  the  value  of  H  from  Eq.  (u),  we  obtain  the  equa- 
tion of  the  parabola, 

x2 


If  the  origin  of  coordinates  is  taken  at  one  of  the  supports 
(as  Aj  Fig.  i),  the  equation  becomes, 


(14) 


The  maximum  tension  in  the  cable,  occurring  at  either  sup- 
port, will  be 


THE  CABLE  5 

or,  by  Eq.  (n), 

wl2  / 7—5-  (    x 

Ti=  —  Vi  +  i6n2, (15) 

°J 

where  n  denotes  the  ratio  of  the  sag/  to  the  span  /: 


Equation  (15)  may  also  be  derived  from  Eq.  (5)  by  noting  that 
the  inclination  of  a  parabolic  cable  at  the  support  is  given  by, 

tan</>i=^  =  4tt.     .      .     .      .     .     (17) 

To  find  the  length  of  the  cable,  Z,,  use  the  general  formula, 


Substituting  the  value  of  —  obtained  from  Eq.  (13),  we  have, 


L  = 

which  yields,  upon  integration, 
/ 


-  log,  [4n+(i  +  i6n2)*].      .     (20) 
2  tin 

This  formula  gives  the  exact  length  of  the  parabola  between  two 
ends  at  equal  elevation. 

For  more  expeditious  solution,  when  a  good  table  of  hyper- 
bolic functions  is  available,  Eq.  (20)  may  be  written  in  the  form 

L  =  -—  (2«+sinh  2w), (20') 

IOW 

where  u  is  defined  by  sinh  u  =  qn. 

An  approximate  formula  for  the  length  of  curve  may  be 
obtained  by  expanding  the  binomial  in  Eq.  (19)  and  then  inte- 
grating. This  gives, 

.  •  •  ),       ...     (21) 


6 


STRESSES  IN   SUSPENSION  BRIDGES 


where  n  is  defined  by  Eq.  (16).     For  small  values  of  the  sag- 
ratio  n,  it  will  be  sufficiently  accurate  to  write, 

Z,  =  /(i+frc2),       .     .     .     .     .     .     (22) 

for  the  length  of  a  parabolic  cable  in  terms  of  its  chord  /. 

The  following  table  gives  the  values  of  L  as  computed  by 
Eqs.  (20)  and  (21),  respectively. 


,  •- 

L 

Length-Ratio  =  - 

Sag-Ratio  ("  =  -} 

' 

Exact 

Approximate 

(Eq.  20) 

(Eq.  21) 

•05 

.00662 

.00663 

•075 

•01475 

.01480 

.1 

.02603 

.02603 

•125 

.04019 

.04010 

.15 

•05693 

.05676 

•175 

.07647 

•07566 

.2 

.09822 

1.09643 

3.  Unsymmetrical  Spans. — If  the  two  ends  of  a  cable  span 

are  not  at  the  same  ele- 
vation, the  ordinates  y 
should  be  measured  ver- 
tically from  the  inclined 
closing  chord  AB  (Fig.  2). 
If  that  is  done,  all  of  the 
principles  derived  above 
will  remain  applicable, 
and  Eqs.  (i)  to  (14),  in- 
clusive, may  be  kept  un- 
changed. For  a  load  uni- 
form along  the  horizontal,  the  curve  will  be  a  parabola,  and  its 
equation,  referred  to  the  origin  A  and  to  the  axis  AB,  will  be 
as  before, 

y»4&(/-*).  (14) 


A       yasKr;*— i 


FIG.  2.-Unsymmetrical  Parabolic  Cable. 


THE  CABLE 


If  it  is  desired  to  refer  the  curve  to  the  horizontal  line  AD, 
with  which  the  closing  chord  makes  an  angle  a,  the  equation 
becomes, 

/  =  >>+£•  tan  a  =  4^(/— x)+x-t&na.  .     .     .     (23) 

To  find  the  lowest  point  in  the  curve,  located  at  F,  a  little  to  one 
side  of  the  center,  differentiate  Eq.  (23)  and  place  the  result 
equal  to  zero.  Solving  for  x,  we  obtain, 


— 7-tanaj (24) 


To  find  the  exact  length  of  the  curve,  apply  Eq.  (20)  to  the 
segments  VA  and  VB  (Fig.  2),  treating  each  of  these  segments 
as  one-half  of  a  complete  parabola,  and  add  the  results. 

An  extreme  case  of  the  unsymmetrical  parabolic  curve  occurs 
in  the  side-span  cables  of  suspension  bridges.  Using  the  nota- 
tion shown  in  Fig.  3,  the  equation  of  the  curve  may  be  written 
in  the  same  way  as  Eq.  (14), 


4/1*1 
/i2 


(25) 


Here,  again,  y\  and  /i  are  measured  vertically  from  the  closing 
chord,  and  x\  and  l\  are 
measured  horizontally. 

The  true  vertex  of  the 
curve  or  lowest  point,  F, 
will  generally  be  found, 
by  an  equation  similar  to 
Eq.  (24),  to  be  outside 
point  D  (Fig.  3).  The 
exact  length  of  curve  will 
be  VA  -  VD,  or  the  differ- 
ence between  two  semi- 
parabolas  each  of  which 
may  be  calculated  by  Eq.  (20). 

An  approximate  value  of  the  length  may  be  obtained  by 


I 4 , 

FIG.  3. — Parabolic  Cable  in  Side  Span. 


STRESSES   IN   SUSPENSION   BRIDGES 


taking  the  closing  chord.  AD  =  li-secai,  and  adding  the  para- 
bolic curvature  correction  as  in  Eq.  (22).     This  will  yield 

T  7    /  ,    8        Wi2      \ 

LI  =  /il  sec  «i+ — I,      ....     (26) 

where 

wi=-^.      .     .     •••     •     •     .     .     .     (27) 

The  cable  tension  in  the  side  span  acts  in  the  line  of  the  closing 
chord  AD  (Fig.  3)  and  is  designated  by 


(28) 


Since  the  lever  arms  y\  are  vertical,  they  must  be  multiplied  by 
the  horizontal  component  of  HI,  or  H,  to  obtain  the  bending 
moments  produced  by  this  force.  Hence,  as  in  Eq.  (2),  we  have, 

M'  =  H-yi,  ...     .    :.      .    |t     (29) 
and,  as  in  Eq.  (n),  we  obtain, 

(      \ 

(30) 


In  order  that  the  main  and  side  spans  may  have  equal  values 
of  H,  by  Eqs.  (n)  and  (30),  we  have, 

Wl2      Will2 


Hence  the  necessary  relation  between  the  sags  is 


The  stress  at  any  point  in  the  cable  is  given  by  Eq.  (5), 
which  may  be  rewritten  as 


(33) 


At  the  center  of  the  side  span,  where  x\  =—  ,  the  curve  is  parallel 


THE   CABLE  9 

to  the  chord,  and  the  inclination  is  equal  to  a\\   hence,  at  that 
point 

!)*  ......  "    (34) 


At  the  support,  where  x\  =  o,  the  inclination  of  the  cable  is  given 
by 

tan  </>!  =  tan  «i  +4^,        .....     (35) 

/i 
and  formula  (33)  yields 


,      .     .     .     (36) 


which  is  the  maximum  stress  in  the  cable. 

4.  The  Catenary.  —  If  the  load  w  is  not  constant  per  hori- 
zontal unit,  but  per  unit  length  of  the  curve,  as  is  the  case  where 
the  load  on  the  cable  is  due  to  its  own  weight,  Eq.  (10)  takes  the 
form, 

d2          w  •  sec  0 


Since  tan  0  =    ^  Eq.  (37)  may  be  written, 

.....    (38) 


Integrating  this  equation,  taking  the  origin  at  the  lowest  point 
of  the  curve,  we  obtain  the  equation  of  the  cable  curve: 

y=±-c(ecx+e-<*-2),  .....     (39) 

w 

where  c  =  —  . 
H 

This  is  the  equation  of  a  catenary;  a  cable  under  its  own 
weight  hangs  in  a  catenary. 

Replacing  the  exponential  terms  by  hyperbolic  functions, 
Eq.  (39)  may  be  written, 

—  i)  .......     (40) 


10  STRESSES   IN   SUSPENSION   BRIDGES 

To  find  the  length  of  the  catenary,  substitute  -j-  obtained 

doc 

from  Eq.  (39)  in  Eq.  (18).     This  gives 

e-cx)dx  =  -(e^-e~^).      .  .  .     (41) 


Expressed  in  hyperbolic  functions,  Eq.  (41)  may  be  written, 

T       2     .    ,    Cl  f      \ 

Z,  =  -smh  —  .......     (42) 

C  2 

Equations  (40)  and  (42)  are  useful  in  computations  for  the 
guide  wires  employed  for  the  regulation  of  the  strands  in  cable 
erection.  If  the  length  L  is  known,  Eq.  (42)  may  be  solved  for 
the  parameter  c,  by  a  method  of  successive  approximations,  and 
the  ordinates  may  then  be  obtained  from  Eq.  (40).  For  the 
expeditious  solution  of  these  equations,  good  tables  of  hyperbolic 
functions  are  required. 

If  the  integration  in  Eq.  (41)  is  performed  between  the  limits 
0  and  x,  and  the  value  of  y  substituted  from  Eq.  (39),  we  obtain, 


(43) 


as  a  formula  for  the  length  from  the  vertex  to  any  point  of  the 
curve.     Equation  (43)  may  be  used  for  unsymmetrical  catenaries. 
The  stress  at  any  point  in  the  cable  is  again  given  by  Eq.  (5), 
or, 

>  -  r-*.|    ._.     .     .     .     .     (44) 

Since  H  =  -,  Eq.  (44)  may  be  written: 


"HDT 


Substituting  the  value  of  -~  derived  from  Eq.  (39),  we  obtain, 


(45) 


THE  CABLE  11 

Replacing  the  exponential  by  hyperbolic  functions,   Eq.   (45) 
becomes, 

T  =  H-coshcx.     .    '-..    .    '  f      .      .     (46) 
This  tension  will  be  a  maximum  at  the  ends  of  the  span,  where 


-,  yielding, 


-.      .      .      .      .      .      .     (47) 


Comparing  Eqs.  (40)  and  (46),  we  find, 

T  =  w(y+-\=wy+H     .      ...     (48) 

At  the  span  center,  where  y  =  o,  this  gives  T  =  H\    and  at  the 
supports,  where  y  =/,  we  obtain, 

.,-     .     .r    .     .     ,     .     (49) 


If  the  sag-rat'o  (^  =  7)  is  small,  all  of  the  formulas  for  the 

V       */ 
catenary  may  be  replaced,   with  sufficient  accuracy,   by  the 

formulas  for  parabolic  cables. 

5.  Deformations  of  the  Cable.  —  As  a  result  of  elastic  elon- 
gation, slipping  in  the  saddles,  or  temperature  changes,  the 
length  of  cable  between  supports  may  alter  by  an  amount  AZ,; 
as  a  result  of  tower  deflection  or  saddle  displacement,  the  span 
may  alter  by  an  amount  A/.  Required  to  find  the  resulting 
changes  in  cable-sag,  A/. 

For  parabolic  cables,  the  length  is  given  with  sufficient 
accuracy  by  Eq.  (21).  Partial  differentiation  of  that  equation 
with  respect  to  /  and  /,  respectively,  yields  the  two  relations  : 

8w4)-A/,       .      .-    .     (50) 
/.      .      .      ....     (51) 

From  Eqs.  (50)  and  (51),  there  results, 

it        15  —  4ow2+288w4    ., 
A=-  'A 


12  STRESSES  IN  SUSPENSION  BRIDGES 

The  required  center  deflections  may  be  calculated  by  means  of 
Eqs.  (51)  and  (52)  when  AL  and  A/  are  known. 

For  a  change  in  temperature  of  /  degrees,  coefficient  of  expan- 
sion a),  the  change  in  cable-length  will  be, 

AZ  =  to-/-L.        .....     (53) 

For  any  loading  which  produces  a  horizontal  tension  H,  the 
average  stress  in  the  cable  will  be,  very  closely, 

L  jj 

rs> 


and  the  elastic  elongation  will  be, 

L     HL 


(54) 


where  E  is  the  coefficient  of  elasticity  and  A  is  the  area  of  cross- 
section  of  the  cable. 

Another  expression  for  the  elastic  elongation  is 

H   CL  ds2    HI 


For  a  small  change  in  the  cable-sag  A/,  the  resulting  change 
in  the  horizontal  tension  is  obtained  by  differentiating  Eq.  (12): 


From  Eqs.  (56),  (51)  and  (52),  may  be  found  the  deformations 
of  the  cable  produced  by  any  small  change  in  the  cable  stresses. 

SECTION  IL—  UNST.IFFENED  SUSPENSION  BRIDGES 

6.  Introduction.  —  The  unstiffened  suspension  bridge  is  not 
used  for  important  structures.  The  usual  form,  as  indicated  in 
Fig.  4,  consists  of  a  cable  passing  over  two  towers  and  anchored 
by  back-stays  to  a  firm  foundation.  The  roadway  is  suspended 
from  the  cable  by  means  of  hangers  or  suspenders.  As  there  is 
no  stiffening  truss,  the  cable  is  free  to  assume  the  equilibrium 
curve  of  the  applied  loading. 


UNSTIFFENED  SUSPENSION  BRIDGES  13 

7.  Stresses  in  the  Cables  and  Towers. — If  built-up  chains 
are  used,  as  in  the  early  suspension  bridges,  the  cross-section 
may  be  varied  in  proportion  to  the  stresses  under  maximum 
loading.  In  a  wire  cable,  the  cross-section  is  uniform  through- 
out. 

As  the  cable  and  hangers  are  light  in  comparison  with  the 
roadway,  the  combined  weight  of  the  three  may  be  considered 
as  uniformly  distributed  along  the  horizontal.  Let  this  total 
dead  load  be  w  pounds  per  lineal  foot.  The  cable  will  then 
assume  a  parabolic  curve;  and  all  of  the  relations  derived  for  a 
parabolic  cable,  represented  by  Eqs.  (n)  to  (22),  will  apply. 


--S,  **r 


FIG.  4. — Unstiffened  Suspension  Bridge. 

The  maximum  dead-load  stress  in  the  cable,  occurring  at  the 
towers,  is  given  by  Eq.  (15) : 


where  n  is  the  ratio  of  the  sag  /  to  the  span  /. 

Let  there  be  a  uniform  live  load  of  p  pounds  per  lineal  foot. 
The  maximum  cable  stress  will  evidently  occur  when  the  load 
covers  the  whole  span,  and  will  have  a  value, 


(57) 


Adding  the  values  in  Eqs.  (15)  and  (57),  we  find  the  total  stress 
in  the  cable  at  the  towers: 


w2)K  ...  ;.  •;   (S8) 


14  STRESSES   IN  SUSPENSION  BRIDGES 

If  ai  is  the  inclination  of  the  backstay  to  the  horizontal 
(Fig.  4),  the  stress  in  the  backstay  will  be: 

ec*:    .     .     (59) 


If  cable  and  backstay  have  equal  inclinations  at  the  tower, 
their  stresses,  represented  by  Eqs.  (58)  and  (59),  will  be  equal. 

The  vertical  reaction  of  the  main  cable  at  the  tower  is 
(w+p)l/2.  If  the  backstay  has  the  same  inclination  as  the 
cable,  it  will  also  have  the  same  vertical  reaction;  so  that  the 
total  stress  in  the  tower  will  be, 

T=(w+p)-l.         .      .      .     .     ,.'    (60) 

8.  Deformations  under  Central  Loading.  —  Under  partial 
loading,  the  unstiffened  cable  will  be  distorted  from  its  initial 

parabolic  curve.  It  is  re- 
quired to  find  the  deflections 
produced  by  the  change  of 
curve,  disregarding  for  the 
present  any  stretching  of  the 
cable  or  any  displacement  of 

FIG.  5.  —  Loading  for  Maximum  Vertical     . 

Deflection.  the  saddles. 

The  maximum  vertical  de- 

flection at  the  center  of  the  cable  will  occur  when  a  certain 
central  portion  of  length  kl  is  covered  with  live  load  (/>),  in 
addition  to  the  dead  load  (w)  covering  the  whole  span  (Fig.  5). 
The  sag  of  the  distorted  cable  will  be,  by  Eq.  (3), 

•  •      r-S+S-*-*   •  •  •  •  «*> 

Equating  the  expressions  for  the  cable-length  corresponding 
to  the  initial  and  distorted  conditions,  respectively,  the  lengths 
being  obtained  from  the  approximate  equation  (22),  and  intro- 
ducing the  symbol  q  —  p/w,  we  obtain: 

.     (62) 


UNSTIFFENED   SUSPENSION  BRIDGES 


15 


Solving  this  equation  for  H,  and  substituting  in  Eq.  (61),  there 
results: 

/  =  _  i  +  2qk-qk2 

f 


(     . 


By  differentiating  this  expression  with  respect  to  k,  we  obtain 
the  following  condition  for  a  maximum  value  of  /'  : 

k*(i  +  2q)q+2k*(i-q)q+3k2(i-q)  -46  +  1=0.    .      (64) 

Solving  this  equation  for  k  and  substituting  the  result  in  Eq.  (63), 
we  obtain  the  following  values  for  the  maximum  crown  deflection 


For  q  =  -!-=  o 

w 


i 


6=1.0    '     0.64          0.30          0.28          O.25          0.23          0.21 

A/=   o         .013        .022        .028        .045        .067        .oSo/ 
From  this  tabulation  we  may  obtain  the  following  empirical 

values,  sufficiently  accurate  between  the  limits  q  =  —  =  -  to  4: 

w    4 


(65) 
A/=  (o  .  007  +o  .  046*7  —  o  .  oo75<?2)/ 

9.  Deformations  under  Unsymmetrical  Loading.  —  The  great- 
est distortion  of  the  cable 
from  symmetry,  repre- 
sented by  the  maximum 
horizontal  displacement  of 
the  low  point  or  vertex, 
will  be  produced  by  a 
continuous  uniform  load 


FIG.  6.  —  Maximum  Horizontal  Displacement 
of  the  Crown. 


extending  for  some  dis- 
tance  kl  from  the  end  of 
the  span.  (Fig.  6.) 

Applying  the  principle  of  Eq.  (3),  the  lowest  point  of  the 

cable  curve  is  located  by  the  condition  ~j~=05    accordingly, 


16  STRESSES  IN   SUSPENSION    BRIDGES 

with  the  notation  of  Fig.  6,  so  long  as  the  crown  V  is  to  the  left 
of  the  head  of  the  load  (E), 

-(l-2x)^pm  =  o.   .      .    V    .     (66) 
2 

Inspection  of  this  equation  shows  that  x  will  have  its  maximum 
value  when  k  has  its  maximum  value;  that  is,  when  kl  =  l  —  x\ 
in  other  words,  the  greatest  lateral  displacement  occurs  when 
the  head  of  the  moving  load  reaches  the  low  point,  V.  Substi- 
tuting this  value  in  Eq.  (66),  we  obtain: 


-f (67) 

Hence  the  maximum  deviation  of  the  crown  (V)  from  the  center 
of  the  span  (c),  will  be  (Fig.  6), 


w    w  ,    . 

+F 

The  total  sag  of  the  cable  is  practically  invariable  for  all 
ordinary  values  of  p/w.  Consequently,  the  uplift  of  the  cable 
at  the  center  of  the  span  will  amount  to 


We  thus  obtain  the  following  values: 

For  ^=    I  <<  i  *  2  3 

w        4  3  2  3 

£=.028     .036     .051      .086     .105     .134     .167     . 
A/=  .003     .004     .008     .021     .030     .045     .062     .oy6/ 

10.  Deflections   Due   to   Elongation   of   Cable.  —  The   total 
length  of  cable,  including  the  backstays  (Fig.  4),  is,  by  Eq.  (21), 

.'     .      .     (70) 


For  a  change  in  temperature  of  t  degrees,  the  total  elongation 

of  cable  will  be 

.    .      .    V.     .     ,     (71) 


UNSTIFFENED  SUSPENSION  BRIDGES  17 

For  the  elongation  of  the  cable  due  to  elastic  strain,  we  may 
write,  by  Eq.  (55), 

sec2<*1].         .     ,     (72) 


In  addition  there  may  be  a  contribution  to  AZ,  from  yielding  of 
the  anchorages. 

If  the  cable  is  capable  of  slipping  over  the  fixed  saddles,  the 
resulting  deflection  A/  is  obtained  by  substituting  the  above 
values  of  AL  in  Eq.  (51). 

If,  however,  a  displacement  of  the  saddles  or  a  movement  of 
the  tops  of  the  towers  will  occur  before  the  cable  will  slip,  any 
elongation  of  the  backstays  will  alter  the  span  (/),  but  not  the 
length  (L),  of  the  cable  in  the  main  span.  In  that  case,  the 
combined  effects  of  temperature  and  elastic  strain  will  give: 


(73) 

and 

/  777  \ 

A/=  —2  sec  «!•  /w//i  -sec  0:1+-=—  ^-  sec2  «ij  .     .      (74) 

Substituting  these  values  of  AL  and  A/  in  Eqs.  (51)  and  (52), 
respectively,  we  obtain  the  resulting  deflection  (A/)  of  the  main 
cable: 

.,  15  Ar     15  —  4ow2+288w4   A  ,     . 

-  ^     '     (7S) 

If  a  displacement  of  the  saddles  (A/)  is  accompanied  by  a 
slipping  of  the  cable,  so  that  the  total  length  of  the  latter  between 
anchorages  (Fig.  4)  remains  unchanged,  then  the  changes  in 
length  and  span  of  the  main  cable  must  satisfy  the  relation 

AL  =  A/-cosai  ......      (76) 

Substituting  these  values  in  Eq.    (75),   the  crown  deflection 
becomes, 


1  5  •  cos  <*i  -  (i  5  -  4on2  +  288»4)     .  (    . 

A/=  16(511  -8  '     '     (77) 


18  STRESSES   IN  SUSPENSION  BRIDGES 

SECTION  III.— STIFFENED  SUSPENSION  BRIDGES 

11.  Introduction. — In  order  to  restrict  the  static  distortions 
of  the  flexible  cable  discussed  in  the  preceding  pages,  there  is 
introduced  a  stiffening  truss  connected  to  the  cable  by  hangers 
(Figs.  7,  15,  16).     The  side  spans  may  likewise  be  suspended 
from  the  cable  (Figs.  10,  n,  18),  or  they  may  be  independently 
supported;  in   the  latter   case   the  backstays  will  be  straight 
(Figs.  15,  1 6,  20).     The  main-span  truss  may  be  simply  sup- 
ported at  the  towers  (Figs,  n,  16),  or  it  may  be  built  continuous 
with  the  side  spans  (Figs.  18,  20).     A  hinge  may  be  introduced 
at  the  center  of  the  stiffening  truss  in  order  to  make  the  struc- 
ture statically  determinate  (Fig.  8),  or  to  reduce  the  degree  of 
indeterminateness. 

Another  form  of  stiffened  suspension  bridge  is  the  braced- 
chain  type.  This  type  does  not  make  use  of  the  straight  stiffen- 
ing truss  suspended  from  a  cable;  instead,  the  suspension  system 
itself  is  made  rigid  enough  to  resist  distortion,  being  built  in  the 
jforrn  of  an  inverted  arch  (Figs.  21,  22,  23,  24). 

For  ease  of  designation,  it  will  be  convenient  to  adopt  a  sym- 
bolic classification  of  stiffened  suspension  bridges,  based  on  the 
number  of  hinges  in  the  main  span  of  the  truss,  as  tabulated  on 
page  19. 

In  types  2F  and  3F,  the  side  spans  are  not  related  to  the  main 
elements  of  the  structure  and  may  therefore  be  omitted  from  con- 
sideration. Hence  these  types  are  called  " single-span  bridges." 

The  suspension  bridges  with  straight  stiffening  trusses  will 
be  analyzed  first. 

12.  Assumptions   Used. — In   the   theory   that   follows,    we 
adopt  the  assumption  that  the  truss  is  sufficiently  stiff  to  render 
the  deformations  of  the  cable  due  to  moving  load  practically 
negligible;    in  other  words,  we  assume,  as  in  all  other  rigid 
structures,  that  the  lever  arms  of  the  applied  forces  are  not 
altered   by   the   deformations   of   the   system.     The  resulting 
theory  is  the  one  ordinarily  employed,  and  is  sufficiently  accu- 
rate for  all  practical  purposes;    any  errors  are  generally  small 
and  on  the  side  of  safety. 


STIFFENED   SUSPENSION  BRIDGES 


19 


Stiffened 

Suspension 

Bridges 


Stiffening  truss 


Continous 

Side  span  free            =  OF 
(Fig.  20) 
Side  span  suspended  =05 
(Fig.  18) 

~      ,  .       ,      /  Side  span  free            =  IF 
One-hinged      < 
I  bide  span  suspended  =15 

Two-hinged     • 

Side  span  free            =2F 
(Fig.  16) 
Side  span  suspended  =  25 
(Fig.  n) 

Three-hinged  • 

Side  span  free            =  3F 
(Fig.  8) 
Side  span  suspended  =  35 
(Fig.  26) 

Braced  chain 


Hingeless 

(Fig.  24) 
One-hinged 
Two-hinged 
(Fig.  23) 
Three-hinged 
(Figs.  21,  22) 


=  \B 
=  2B 

=  3B 


If  the  stiffening  truss  is  not  very  stiff  or  if  the  span,  is  long, 
the  deflections  of  truss  and  cable  may  be  too  large  to  neglect. 
To  provide  for  such  cases,  there  has  been  developed  an  exact 
method  of  calculation  which  takes  into  account  the  deformations 
of  the  system.  For  lack  of  space,  this  "  Exact  Theory"  will 
not  be  presented  here,  but  the  interested  reader  is  referred 
instead  to  other  works  on  the  subject.* 

The  common  theory  developed  in  the  following  pages  for  the 
analysis  of  suspension  bridges  with  stiffening  trusses  is  based  on 
five  assumptions,  which  are  very  near  the  actual  conditions  : 

1.  The  cable  is  supposed  perfectly  flexible,  freely  assuming 
the  form  of  the  equilibrium  polygon  of  the  suspender  forces. 

2.  The  truss  is  considered  a  beam,  initially  straight  and 

*MELAN-STEINMAN:  "Theory  of  Arches  and  Suspension  Bridges,"  pp.  76-86. 
McGraw-Hill  Book  Co.  1913. 

JOHNSON,  BRYAN  AND  TURNEAURE:  "Modern  Framed  Structures,"  Part  II, 
pp.  276-318.  *  John  Wiley  &  Sons.  1911. 

BURR,  W.  H.  :  "  Suspension  Bridges,"  pp.  212-247.    John  Wiley  &  Sons.     1913. 


20 


STRESSES   IN   SUSPENSION  BRIDGES 


horizontal,  of  constant  moment  of  inertia  and  tied  to  the  cable 
throughout  its  length. 

3.  The  dead  load  of  truss  and  cable  is  assumed  uniform  per 
lineal  unit,  so  that  the  initial  curve  of  the  cable  is  a  parabola. 


/vvvvvvvvv 


V 


*r ->^P 


FIG.  7. — Forces  Acting  on  the  Stiffening  Truss. 

4.  The  form  and  ordinates  of  the  cable  curve  are  assumed  to 
remain  unaltered  upon  application  of  loading. 

5.  The  dead  load  is  carried  wholly  by  the  cable  and  causes 


STIFFENED  SUSPENSION  BRIDGES  21 

no  stress  in  the  stiffening  truss.  The  truss  is  stressed  only  by 
live  load  and  by  changes  of  temperature. 

The  last  assumption  is  based  on  erection  adjustments,  involv- 
ing regulation  of  the  hangers  and  riveting-up  of  the  trusses  when 
assumed  conditions  of  dead  load  and  temperature  are  realized. 

13.  Fundamental  Relations.  —  Since  the  cable  in  the  stiffened 
suspension  bridge  is  assumed  to  be  parabolic,  the  loads  acting 
on  it  must  always  be  uniform  per  horizontal  unit  of  length.  All 
of  the  relations  established  for  a  uniformly  loaded  cable  (Eqs.  (n) 
to  (36),  inclusive)  will  apply  in  this  case. 

If  the  panel  points  are  uniformly  spaced  (horizontally),  the 
suspender  forces  must  be  uniform  throughout  (Fig.  7).  These 
suspender  forces  are  loads  acting  downward  on  the  cable,  and 
upward  on  the  stiffening  truss.  It  is  the  function  of  the  stiffen- 
ing truss  to  take  any  live  load  that  may  be  arbitrarily  placed 
upon  it  and  distribute  it  uniformly  to  the  hangers. 

The  cable  maintains  equilibrium  between  the  horizontal  ten- 
sion H  (resisted  by  the  anchorages)  and  the  downward  acting 
suspender  forces.  If  these  suspender  forces  per  horizontal  linear 
unit  are  denoted  by  s,  they  are  given  by  Eq.  (n)  as 


The  truss  (Fig.  7)  must  remain  in  equilibrium  under  the 
arbitrarily  applied  loads  acting  downward  and  the  uniformly 
distributed  suspender  forces  acting  upward.  If  we  imagine  the 
latter  forces  removed,  then  the  bending  moment  M'  and  the 
shear  V  at  any  section  of  the  truss,  distant  x  from  the  left  end, 
may  be  determined  exactly  as  for  an  ordinary  beam  (simple  or 
continuous  according  as  the  truss  rests  on  two  or  more  supports)  . 
This  moment  and  shear  would  be  produced  if  the  cable  did  not 
exist  and  the  entire  load  were  carried  by  the  truss  alone.  If 
—  M8  represents  the  bending  moment  of  the  suspender  forces  at 
the  section  considered,  then  the  total  moment  in  the  stiffening 
truss  will  be 

M=M'-M..       ....     .     (79) 


22  STRESSES  IN   SUSPENSION  BRIDGES 

Similarly,  if  —  Vs  represents  the  shear  produced  by  the  suspender 
forces  at  the  same  section,  the  total  shear  in  the  stiffening  truss 
will  be 

V=V'-V.  .....      .      .     (80) 

Equations  (79)  and  (80)  are  the  fundamental  formulas  for 
determining  the  stresses  in  any  stiffening  truss.  By  these 
formulas,  the  stresses  can  be  calculated  for  any  given  loading 
as  soon  as  the  value  of  H  is  known. 

The  dead  load  is  assumed  to  be  exactly  balanced  by  the 
initial  suspender  forces,  so  that  it  may  be  omitted  from  considera- 
tion in  these  equations. 

In  calculating  M'  and  V  from  the  specified  live  load,  and 
Ms  and  Vs  from  the  uniform  suspender  loading  given  by  Eq.  (78), 
the  condition  of  the  stiffening  truss  as  simple  or  continuous  must 
be  taken  into  account. 

If  the  stiffening  truss  is  a  simple  beam  (hinged  at  the  towers), 
by  a  familiar  property  of  the  funicular  polygon,  represented  by 

Eq.  (2), 

Ms  =  H.y,    .     ,      .    -.      .    ..     (81) 

where  y  is  the  ordinate  to  the  cable  curve  measured  from  the 
straight  line  joining  A'  and  B'  ',  the  points  of  the  cable  directly 
above  the  ends  of  the  truss  (Fig.  7).  Consequently,  Eq.  (79) 
may  be  written, 

M  =  M'-H-y.       .     .      .      .    ...-.    (82) 

which  is  identical  with  equation  (i).  (In  the  unstiffened 
suspension  bridge,  M  =  o.) 

If  4>  is  the  inclination  of  the  cable  at  the  section  considered, 
the  shear  produced  by  the  hanger  forces  is  given  by  Eq.  (7)  as, 


.....    v     (83) 
Consequently,  Eq.  (80)  may  be  written 

F=F'-#-tan0  .....     (84) 

If  the  two  ends  of  the  cable,  A'  and  B1  ',  are  at  unequal  elevations 
(Fig.  7)  ,  Eq.  (84)  must  be  corrected  to  the  form, 

V  =  V'-H  (tan  0  -  tan  a)  ,  (84') 


STIFFENED  SUSPENSION  BRIDGES  23 

where  a  is  the  inclination  of  the  closing  line  A'E'  below  the 
horizontal. 

In  Eqs.  (82),  (84)  and  (84'),  the  last  term  represents  the 
relief  of  bending  moment  or  shear  by  the  cable  tension  H. 

Representing  Mf  by  the  ordinates  yf  of  an  equilibrium  polygon 
or  curve,  constructed  for  the  applied  loading  with  a  pole  distance 
=  H,   Eq.  (82)  takes  the  form, 
/ 

M  =  H(y'-y) (85) 

Hence  the  bending  moment  at  any  section  of  the  stiffening 
truss  is  represented  by  the  vertical  intercept  between  the  axis 
of  the  cable  and  the  equilibrium  polygon  for  the  applied  loads 
drawn  through  the  points  A'B'  (Fig.  7). 

If  the  stiffening  truss  is  continuous  over  several  spans,  the 
relations  represented  by  Eqs.  (81)  to  (85),  inclusive,  must  be 
modified  to  take  into  account  the  continuity  at  the  towers. 
The  corresponding  formulas  will  be  developed  in  the  section  on 
continuous  stiffening  trusses  (Section  VI). 

14.  Influence  Lines. — To  facilitate  the  study  and  determina- 
tion of  suspension  bridge  stresses  for  various  loadings,  influence 
diagrams  are  most  convenient. 

The  base  for  all  influence  diagrams  is  the  #-curve 
or  ^-influence  line.  This  is  obtained  by  plotting  the  equations 
giving  the  values  of  H  for  varying  positions  of  a  unit  concentra- 
tion. In  the  case  of  three-hinged  suspension  bridges,  the  ^-influ- 
ence line  is  a  triangle  (Figs.  8  and  9) .  In  the  case  of  two-hinged 
stiffening  trusses,  the  /7-lines  (Figs,  n,  14)  are  similar  to  the 
deflection  curves  of  simple  beams  under  uniformly  distributed 
load.  In  the  case  of  continuous  stiffening  trusses,  the  #-line 
(Fig.  1 8)  is  similar  to  the  deflection  curve  of  a  three-span  con- 
tinuous beam  covered  with  uniform  load  in  the  suspended 
spans. 

To  obtain  the  influence  diagrams  for  bending  moments  and 
shears,  all  that  is  necessary  is  to  superimpose  on  the  ZT-curve, 
as  a  base,  appropriately  scaled  influence  lines  for  moments  and 
shears  in  straight  beams. 


24  STRESSES  IN  SUSPENSION  BRIDGES 

The  general  expression  for  bending  moments  at  any  section 
(Eq.  82)  may  be  written  in  the  form, 


(86) 


(excepting  that  in  the  case  of  continuous  stiffening  trusses,  y  is 
to  be  replaced  by  y  —  ef\  see  Eq.  212).  For  a  moving  con- 

centration, —  represents  the  moment  influence  line  of  a  straight 

beam,  simple  or  continuous  as  the  case  may  be,  constructed 
with  the  pole  distance  y.  Hence  the  moment  M  is  proportional 
to  the  difference  between  the  ordinates  of  this  influence  line  and 
those  of  the  ^-influence  line.  If  the  two  influence  lines  are 
superimposed  (Figs.  Sb,  nb,  nc,  iSb),  the  intercepts  between 
them  will  represent  the  desired  bending  moment  M.  In  the 
case  of  stiffening  trusses  with  hinges  at  the  towers,  M'  is  the 
same  as  the  simple-beam  bending  moment,  and  its  influence 
line  is  familiarly  obtained  as  a  triangle  whose  altitude  at  the 
given  section  is, 


For  a  parabolic  cable,  this  reduces  (by  Eq.  14)  to 

.        *-'      .     .     /'.     .     .     .     (88) 

y    4/ 

K/[f 
Hence  the  —  triangles  for  all  sections  will  have  the  same  altitude 

y 

—  (Figs.  86,  nb).    The  corresponding  altitude  for  sections  in 

/ 
the  side  spans  is  -J-  (Fig.  nc).    The  areas  intercepted  between 

4/i 
the  #-line  and  the  —  triangles,  multiplied  by  py,  give  the  maxi- 

mum and  minimum  bending  moments  at  the  given  section,  X, 
of  the  stiffening  truss.  Areas  below  the  F-line  represent  posi- 
tive moments,  and  those  above  represent  negative  moments 


STIFFENED  SUSPENSION  BRIDGES  25 

(Figs.  8,  n,  18).  Where  the  two  superimposed  lines  intersect, 
we  have  a  point  K,  which  may  be  called  the  zero  point,  since  a 
concentration  placed  at  K  produces  zero  bending  stress  at  X. 
K  is  also  called  the  critical  point,  since  it  determines  the  limit 
of  loading  for  maximum  positive  or  negative  moment  at  X. 
Load  to  one  side  of  K  yields  plus  bending,  and  load  to  the  other 
side  produces  negative  bending. 

The  shear  at  any  section  of  the  stiffening  truss  is  given  by 
Eq.  (84) ,  which  may  be  written  in  the  form, 


(If  the  two  ends  of  the  cable  span  are  at  different  elevations, 
tan  0  in  this  equation  is  to  be  replaced  by  tan  <f>  —  tan  a,  where 
a  is  the  inclination  of  the  closing  chord  below  the  horizontal. 
See  Eq.  84').  For  any  given  section  X,  tan  </>,  the  slope  of  the 
cable,  is  a  constant  and  is  given  by, 

....  (90) 

The  values  assumed  by  the  bracketed  expression  in  Eq.  (89)  for 
different  positions  of  a  concentrated  load  may  be  represented  as 
the  difference  between  the  ordinates  of  the  #-line  and  those  of 
the  influence  line  for  the  shears  V ',  the  latter  being  reduced 

in  the  ratio .     The  latter  influence  line  is  familiarly  obtained 


by  drawing  the  two  parallel  lines  as  and  U  (Figs,  ga,  gb,  140), 
their  direction  being  fixed  by  the  end  intercepts 

.     .     (01) 


tan</>-tano: 


The  vertices  5  and  /  lie  on  the  vertical  passing  through  the  given 
section  X.  The  maximum  shears  produced  by  a  uniformly 
distributed  load  are  determined  by  the  areas  included  between 
the  H  and  V  influence  lines;  all  areas  below  the  #-line  are  to  be 
considered  positive,  and  all  above  negative.  These  areas  must 


26  STRESSES  IN  SUSPENSION   BRIDGES 

be  multiplied  by  />-tan  0  (or  by  ^[tan  0  —  tana])  to  obtain  the 
greatest  shear  V  at  the  section;  and  V  must  be  multiplied  by  the 
secant  of  inclination  to  get  the  greatest  stress  in  the  web  members 
cut  by  the  section. 

SECTION  IV.—  THREE-HINGED  STIFFENING  TRUSSES 

15.  Analysis.  —  This  is  the  only  type  of  stiffened  suspension 
bridge  that  is  statically  determinate  (Types  3F,  35,  35).  The 
provision  of  the  stiffening  truss  with  a  central  hinge  furnishes  a 
condition  which  enables  H  to  be  directly  determined;  viz.,  at 
the  section  through  the  hinge  the  moment  M  must  equal  zero. 
Consequently,  if  the  bending  moment  at  the  same  section  of  a 
simple  beam  is  denoted  by  M'o,  and  if  /  is  the  ordinate  of  the 
corresponding  point  of  the  cable,  by  Eq.  (82), 


Hence  the  value  of  H  for  any  loading  is  equal  to  the  simple- 
beam  bending  moment  at  the  center  hinge  divided  by  the  sag  /. 
Accordingly,  the  cable  will  receive  its  maximum  stress  when  the 
full  span  is  covered  with  the  live  load  p.  In  that  case  Eq.  (92) 
yields 

B-Tf>      ......     (93) 

and,  comparing  this  with  Eq.  (78),  we  see  that 

s  =  p.    ...  ...     (94) 

Hence,  under  full  live  load,  the  conditions  are  similar  to  those 
for  dead  load,  the  cable  carrying  all  the  load,  the  trusses  having 
no  stress.  The  bending  moment  at  any  section  will  be 

Total  M  =  o.      .     .     .     .     .      (94') 

For  a  single  load  P  at  a  distance  kl  from  the  near  end  of  the 
span,  the  simple-beam  moment  at  the  center  hinge  will  be 

PbJ 

if'o~.       .    _.     ••,...    -     (95) 


THREE-HINGED   STIFFENING  TRUSSES  27 

Hence,  the  value  of  H,  by  Eq.  (92),  will  be 

.  -   .    -  B-™,      .,'.'    •    V,    •     •     (96) 

This  value  of  H  will  be  a  maximum  for  k  =  |,  yielding, 


=     ........     (97) 

According  to  Eq.  (92),  the  influence  line  for  H  will  be  similar 
to  the  influence  line  for  bending  moment  M  'o  at  the  center  of  a 
simple  beam;  hence  it  will  be  a  triangle.  It  is  defined  by  Eq. 
(96);  and  its  maximum  ordinate  (at  the  center  of  the  span)  is 
given  by  Eq.  (97)  as  //4/.  Figures  Sb  and  ga  show  the  ^-influence 
line  constructed  in  this  manner. 

If  the  truss  is  uniformly  loaded  for  a  distance  kl  from  one 
end,  the  value  of  H  may  be  found  by  integrating  Eq.  (96)  or 
directly  from  Eq.  (92).  We  thus  obtain: 

for£<i    £r  =  ^.(#),      .....     (98) 
47 

for£>J,     H  =  ^(4k-2k*-i).    .      .      .     (99) 
°7 

For  full  load  (k  —  i),  Eq.  (99)  gives  the  maximum  value  of  //: 

-f  .......  <•»> 

which  is  identical  with  Eq.  (93).  Equations  (98)  to  (100),  inclu- 
sive, may  also  be  obtained  directly  from  the  //-influence  line 
(Figs.  Sb  and  90). 

For  the  half-span  loaded,  Eqs.  (98)  and  (99)  yield, 


which  is  one-half  of  the  value  for  full  load.     Substituting  this 
value  in  Eq.  (78),  we  find, 

s  =  $P  .....  .     -     *     (102) 

One-half  of  the  span  is  thus  subjected  to  an  unbalanced  upward 
load,  s  =  %p,  per  lineal  foot,  and  the  other  half  sustains  an  equal 


28  STRESSES  IN  SUSPENSION  BRIDGES 

downward  load,  p  —  s  =  \p.  Consequently  there  will  be  produced 
positive  moments  in  the  loaded  half,  and  equal  negative  moments 
in  the  unloaded  half,  amounting  to 

M  =  $px(--x\;   .     .     .     .     .     (103) 

and  the  maximum  moments  for  this  loading,  occurring  at  the 
quarter  points,  (#  =  J/,  #  =  !/),  will  be, 

."     ,     .     (104) 


FIG.  8. — Three-hinged  Stiffening  Truss  Moment  Diagrams. 
(Type  3/0. 

16.  Moments  in  the  Stiffening  Truss. — The  influence  dia- 
grams for  bending  moments  are  constructed,  in  accordance  with 

Eq.  (86) ,  by  superimposing  the  —  triangles  upon  the  ^-influence 

y 

triangle.     By  Eq.  (88) ,  the  —  triangles  for  all  sections  have  the 

same  altitude  —  ;  and,  in  the  case  of  the  three-hinged  stiffening 
4/ 


THREE-HINGED   STIFFENING  TRUSSES  29 

truss,  this  altitude  is  identical  with  that  of  the  ^-influence 
triangle. 

The  two  triangles  are  shown  superimposed  in  Fig.  Sb.  The 
shaded  area  between  them  is  the  influence  diagram  for  bending 
moment  at  the  section  X. 

For  x  =  ~,   the  two  triangles  would  coincide.     Hence  the 
2 

moment  at  the  center  hinge  is  zero  for  all  conditions  of  loading, 
which  agrees  with  the  condition  that  the  hinge  can  carry  no 
bending. 

For  x  <-,  the  two  influence  triangles  intersect  at  a  point  K,  a 

short  distance  to  the  left  of  the  center.  K  is  the  zero  point  or 
critical  point.  All  load  to  the  left  of  K  yields  plus  bending,  and 
all  load  to  the  right  produces  negative  bending. 

Since  the  two  superimposed  triangles  have  the  same  base  and 
equal  altitudes,  the  plus  and  minus  intercepted  areas  will  be 
equal.  Hence,  if  the  whole  span  is  loaded,  the  two  areas  will 
cancel  each  other,  yielding  zero  moment  as  required  by  Eq.  (94'). 

If  either  of  the  shaded  areas  is  multiplied  by  py,  it  will  give 
the  maximum  value  of  the  bending  moment  at  X.  The  bending 
moments  may  also  be  obtained  analytically  from  Eq.  (82),  as 
follows  : 

If  the  load  covers  a  length  kl  from  one  end  of  the  span,  the 
bending  moment  at  any  section  x<kl,  by  Eqs.  (82),  (98)  and 
(14),  will  be, 

i-2k2).        .     .     (105) 


Setting  —  -  =  o  in  this  equation,  we  find  that  for  maximum  M, 
dk 


(106) 


This  equation  defines  the  distance  kl  to  the  critical  point  K 
(Fig.  8&).  For  this  value  of  k,  Eq.  (105)  gives  the  maximum 
value  of  M  for  any  value  of  x: 

TIT  HJT      px(l  —  X)(l—2X)  ,         ^ 

Max.  M  =  ^      ,  /      .  —  -.      .     .     .     (107) 


30  STRESSES  IN  SUSPENSION  BRIDGES 


This  value  may  also  be  obtained  from  the  shaded  areas  in 

the  influence  diagram   (Fig.   86).     Setting   —  —  =  o  in  the  last 

ax 

equation,  we  find  that  the  absolute  maximum  M  occurs  at, 


(108) 

Substituting  this  value  in  Eqs.  (107)  and  (106),  we  find  that  the 
absolute  maximum  value  of  M  is, 

Abs.  Max.  lf=+o.oi883/>/2,    .      .      .     (109) 

or  about  -fopl2,  and  that  it  occurs  at  x  =  0.2341,  when  £  =  0.395. 
By  loading  the  remainder  of  the  span  (o  .  6o5/)  ,  we  obtain  the 
maximum  negative  moment  at  the  same  section.  This  will  be 
numerically  equal  to  the  maximum  positive  moment,  since  their 
summation  at  any  section  must  give  zero  according  to  Eq.  (940. 
Hence  the  absolute  maximum  negative  moment  will  be, 

Abs.  Min.  M=  -0.01883^/2.     •      •      •    (109') 

After  the  maximum  moments  at  the  different  sections  along 
the  span  are  evaluated  from  the  influence  lines,  or  from  Eq.  (107), 
they  may  be  plotted  in  the  form  of  curves,  as  shown  in  Fig.  Sc. 
For  the  three-hinged  stiffening  truss,  these  maximum  moment 
curves  are  symmetrical  about  the  horizontal  axis.  They  may 
be  used  as  a  guide  for  proportioning  the  chord  sections  of  the 
stiffening  truss. 

17.  Shears  in  the  Stiffening  Truss.  —  The  shears  produced 
in  the  stiffening  truss  by  any  loading  are  given  by  Eq.  (84)  ; 
but  the  maximum  values  at  the  different  sections  are  most  con- 
veniently determined  with  the  aid  of  influence  lines  (Fig.  9). 

The  influence  line  for  H  is  a  triangle,  with  altitude  =  —  at 

47 

the  center  of  the  span.  Upon  this  is  superimposed  the  influence 
line  for  shears  in  a  simple  beam,  reduced  in  the  ratio  i  :  tan  <j>. 
The  resulting  influence  diagram  for  shear  V  at  a  given  section 

x  <-  is  shown  in  Fig.  ga.     There  are  two  zero  points  or  critical 

4 
points:   at  x  and  at  kl.    The  portion  of  the  left  span  between 


THREE-HINGED    STIFFENING   TRUSSES 


31 


these  two  points  must  be  loaded  to  produce  maximum  positive 
shear  at  the  given  section.  From  the  geometry  of  the  figure 
we  find  the  vosition  of  the  critical  point  K  to  be  given  by, 


k  = 


(no) 


\  fJ—JryfaenOff  —Lin 

^  Inf/uer)ce  -L/ne  /or   l/'-f  /an  7> 


re) 


FIG.  9. — Shear  Diagrams  for  Three-hinged  Stiffening  Truss. 
(Type  3F). 

With  the  load  covering  the  length  from  x  to  kl,  we  find  the 
maximum  positive  shear  at  x,  either  from  the  diagram  or  from 
Eq.  (84),  to  be  given  by 


Max.  F=^- 

2 


32  STRESSES   IN   SUSPENSION  BRIDGES 

When  x  =  o,  or  for  end-shear,  Eq.  (no)  gives  k=%,  and  Eq.  (in) 
yields, 

Abs.  Max.  V=&-.       .     .    ,.     .     (112) 
When  x  =  ^l,  we  find  k  =  %,  and 

Max.  V-£.       .     .     .     .     (113) 

For  x>  J/,  the  influence  diagram  takes  the  form  shown  in  Fig.  gb. 
There  is  only  one  zero  point,  namely  at  the  section  X.  Loading 
all  of  the  span  beyond  X,  we  find  the  maximum  positive  shear, 
either  from  the  diagram  or  from  Eq.  (84),  to  be  given  by, 


Max.  7=3-.      ,     .     .     (114) 


This  has  its  greatest  value  for  x  =  %l,  or  at  the  center,  where  it 
has  the  value, 


Max.  V=. 


Writing  expressions  for  the  maximum  negative  shears  in  the 
same  manner,  we  obtain  values  identical  with  Eqs.  (in)  to 
(115),  but  with  opposite  sign.  In  other  words,  the  plus  and 
minus  areas  in  any  shear  influence  diagram  are  equal;  their 
algebraic  sum  must  be  zero,  since  full  span  loading  produces  no 
stress  in  the  stiffening  truss.  (See  Eq.  94). 

Figure  gc  gives  curves  showing  the  variation  of  maximum 
positive  and  negative  shears  from  end  to  end  of  the  span.  The 
curves  are  a  guide  for  the  proportioning  of  the  web-  members  of 
the  stiffening  truss.  For  the  three-hinged  truss,  these  curves 
are  symmetrical  about  the  horizontal  axis. 

If  the  two  ends  of  the  cable' are  at  unequal  elevations,  the 
foregoing  results  for  shear  (Eqs.  (no)  to  (115),  inclusive)  must 
be  modified  on  account  of  the  necessary  substitution  through- 
out the  analysis  of  (tan  0  — tan  a)  for  tan  <j>  as  required  by 
Eq.  (84')- 


TWO-HINGED  STIFFENING  TRUSSES 


33 


SECTION  V.— TWO-HINGED 
STIFFENING  TRUSSES 

18.  Determination  of  the 
Horizontal    Tension    H. — In 

these  bridge  systems,  the  hori- 
zontal tension  H  is  statically 
indeterminate.  The  required 
equation  for  the  determina- 
tion of  H  must  therefore 
be  deduced  from  the  elas- 
tic deformations  of  the  sys- 
tem. 

Imagine  the  cable  to  be 
cut  at  a  section  close  to  one 
of  the  anchorages.  Then 
(with  H  =  o),  under  the  action 
of  any  loads  applied  on  the 
bridge,  the  two  cut  ends  would 
separate  by  some  horizontal 
distance  A.  If  a  unit  hori- 
zontal force  (H  =  i)  be  applied 
between  the  cut  ends,  it  would 
pull  them  back  toward  each 
other  a  small  distance  5.  The 
total  horizontal  tension  H  re- 
quired to  bring  the  two  ends 
together  again  would  evi- 
dently be  the  ratio  of  these 
two  imaginary  displacements, 
or, 

H  =  ±  (116) 


CX  G 

C/3          .0 


•      a 

& 


Substituting  for  A  and  5 
the  general  expressions  for 
the  displacement  of  a  point 


34  STRESSES  IN  SUSPENSION  BRIDGES 

in  an  elastic  system  under  the  action  of  given  forces,  Eq.  (116) 
becomes, 


.    (117) 


where  M'  =  bending  moments    (in  the  stiffening    truss)   under 

given  loads,  for  H  =  o. 
m  =  bending  moments  (in  the  stiffening  truss)  with  zero 

loading,  f or  H  =  i . 
u  =  direct  stresses   (in  the  cable,  towers  and  hangers) 

with  zero  loading,  f  or  H  =  i . 
I  =  moments  of  inertia  (of  the  stiffening  truss) . 
A  =  areas   of   cross-section    (of   the   cable,    towers   and 
hangers) . 

In  the  denominator  of  Eq.  (117)  there  are  two  terms,  since  the 
system  is  made  up  of  members,  part  of  which  are  acted  upon  by 
bending  moments,  and  part  by  direct,  or  axial,  stresses.  In 
the  numerator,  there  is  no  axial  stress  term,  since  for  the  con- 
dition of  loading  producing  A,  the  cable  tension  H  =  o,  and  all  of 
the  axial  stresses  (in  cables,  towers  and  hangers)  vanish. 

Equation  (117)  is  the  most  general  form  of  the  expression 
for  H,  and  applies  to  any  type  of  stiffened  suspension  bridge. 

When  there  are  no  loads  on  the  span,  the  bending  moments 
in  the  two-hinged  stiffening  truss  are,  by  Eq.  (82) : 

M=-H-y.        .      .     .     .     .     (118) 

Hence  we  have,  when  H  =  i, 

m=-y (119) 

The  stress  at  any  section  of  the  cable  is  given  by  Eq.  (5),  which, 
for  H  =  i,  reduces  to, 

ds  ' 

«=-.  .    ,     .     .   ,.     .     . 


TWO-HINGED  STIFFENING  TRUSSES  35 

Substituting  Eqs.  (119)  and  (120)  in  Eq.  (117),  we  obtain  the 
following  fundamental  equation  for  H  for  two-hinged  stiffening 
trusses  (Types  2F  and  25)  : 


( 

J     EI 


Cfd*    r 

J    EI     J 


f          \ 

'   '   (I2I) 


EAdx2 


The  integral  in  the  numerator  and  the  first  integral  in  the 
denominator  represent  the  contributions  of  the  bending  of  the 
stiffening  truss  to  A  and  5  respectively;  the  integrations  extend 
over  the  full  length  of  the  stiffening  truss  suspended  from  the 
cable.  The  second  integral  in  the  denominator  represents 
the  contribution  of  the  cable  stretch  to  the  value  of  6;  the 
integration  extends  over  the  full  length  of  cable  between 
anchorages. 

In  the  denominator  of  Eq.  (121),  the  truss  term  contributes 
about  95  per  cent,  and  the  cable  term  only  about  5  per  cent  of 
the  total.  Hence,  certain  approximations  are  permissible  in 
evaluating  the  cable  term. 

Terms  for  the  towers  and  hangers  have  been  omitted,  as  they 
are  negligible.  (Their  contribution  to  the  denominator  would 
be  only  a  small  decimal  of  i  per  cent.) 

The  terms  in  the  denominator  are  independent  of  the  loading 
and  will  now  be  evaluated.  See  Fig.  na  for  the  notation 
employed.  The  symbols  /,  x,  y,  /,  a,  I,  A  have  already  been 
defined  for  the  main  span;  and,  adding  a  subscript,  we  have  the 
corresponding  symbols  /i,  xi,  yi,  /i,  «i,  /i,  AI}  for  the  side 
spans.  In  addition  we  have, 

/'  =  span  of  the  cable,  center  to  center  of  towers,  which 

distance  may  be  somewhat  greater  than  the  truss 

span  /  (Fig.  na); 
/2  =  horizontal   distance   from   tower   to   anchorage,   which 

distance  may  be  greater  than  the  truss  span  /i  (Fig. 

na). 

Substituting  for  y  its  values  from  Eqs.  (14)  and  (25),  the 


36 


STRESSES  IN  SUSPENSION  BRIDGES 


first  integral  in  the  denominator  of  Eq.  (121),  extending  over 
main  and  side  spans,  becomes, 

J    Ch  8  /2/  ,     /  8    fi2h\      ,      . 

'      ^+2-  (I22) 


The  moments  of  inertia  /  and  /i  are  here  assumed  constant  over 
the  respective  spans. 

I 


i   u— *-  — I 


-| L ^  j  Juo -yi/nyec/\ Ou 

It M • 


_ia^5^r__  _^  __H'  u 


FIG.  ii.— Moment  Diagrams  for  Two-hinged  Stiffening  Truss.     (Type  25). 

The  second  integral  in  the  denominator  of  Eq.  (121),  with 
the  aid  of  Eqs.  (13)  and  (23),  may  be  written, 


TWO-HINGED   STIFFENING  TRUSSES  37 

The  cable  sections  A  and  A\  are  here  considered  to  be  uniform 
in  the  respective  spans.  Usually  A=A\.  Expanding  the 
binomials  and  integrating,  we  obtain,  with  sufficient  accuracy, 

7~r~2=^~r(I+8w2)+^~T"sec3ai(I+8wi2)>  (I24)* 


where  n  and  n\  are  the  sag-ratios  in  main  and  side  spans,  respect- 
ively : 

n={,    «!=£  .......     (124') 

/  /I 

Setting  the  values  given  by  (122)  and  (124)  in  Eq.  (121), 

*EI 
and  multiplying  through  by  ^T,  the  formula  for  H  becomes, 


J-l"  Clfydx+i  C'Mi' 

E=  -  //IJ°        .    .» 


(125) 


where,  for  abbreviation, 

;  ||    .••-£•  "I:  -f  •  •  •  •  (I26) 

The  elastic  coefficient  EC  =  E,  unless  the  cable  is  made  of  wire 
ropes.  The  denominator  of  Eq.  (125),  to  be  used  for  all  suspen- 
sion bridges  of  Type  25,  will  henceforth  be  designated  by  N. 
It  is  a  constant  for  any  given  structure. 

The  second  term  in  the  numerator  represents  the  contribu- 
tion of  any  loads  in  the  side  spans,  and  will  vanish  if  the  side 
spans  are  built  independent  of  the  backstays.  In  the  latter  case 


*  If  the  cable  section  is  not  constant  but  varies  with  the  cable  stress  (as 
in  eyebar  chains),  change  8w2  to  -^n2,  8w,.2  to  -^MI*,  and  sec3  ai  to  sec2  a\\ 
using  A0  (cable  section  at  crown)  instead  of  A  and  AI. 


38  STRESSES  IN  SUSPENSION  BRIDGES 

the  backstays  will  be  straight  (Type  2F,  Fig.   16),  all  terms 
containing  yi,fi,  n\,  or  v  will  vanish,  and  Eq.  (125)  reduces  to 


H= 


A  fM'ydx 


8     E 


where  a\  is  the  slope  of  the  backstay. 

19.  Values  of    H  for  Special  Cases  of  Loading.  —  In  the 

preceding  equations,  the  value  of  M'  depends  upon  the  loading 
in  the  particular  case.  Expressing  M'  as  a  function  of  x,  using 
the  value  of  y  given  by  Eq.  (14),  and  performing  the  integration 
as  indicated,  we  find,  for  a  single  load  P  at  a  distance  kl  from 
either  end  of  the  span, 


C 


;"•    .     (128) 


Hence,  by  Eq.  (125),  for  a  concentration  in  the  main  span,  the 
value  of  the  horizontal  tension  will  be, 

,        ....     (129) 


where  N  denotes  the  denominator  of  Eq.  (125),  and  the  function 

,        .        .       V  (129') 


and  may  be  obtained  directly  from  Table  I  or  from  the  graph  in 
Fig.  12.  The  above  value  of  H  is  a  maximum  when  the  load  P 
is  at  the  middle  ot  the  span;  then  k=\,  and  Eq.  (129)  yields, 


Similarly,  '  for  a  concentration  P  in  either  side  span,  at  a 
distance  k\l\  from  either  end, 

.P,   .     ..     .    -..     (131) 


*  If  the  cable  section  is  not  constant  but  varies  with  the  cable  stress  (as  in 
eyebar  chains),  change  8w2  to  -^w2,  and  sec3  «i  to  sec2  a\\  using  A0  (cable 
section  at  crown)  instead  of  A  and  A\. 


TWO-HINGED   STIFFENING  TRUSSES 


39 


9(        8,        $        3         °. 


40  STRESSES  IN   SUSPENSION  BRIDGES 

where  B(k\)  is  the  same  function  as  defined  by  Eq.  (129'). 
This  value  of  H  is  a  maximum  when  the  load  P  is  at  the  middle 
of  the  side  span;  then  k\  =  |,  and  Eq.  (131)  yields, 


By  plotting  Eqs.  (129)  and  (131)  for  different  values  of  k 
and  ki,  we  obtain  the  ZZ-curves  or  influence  lines  for  H  (Figs. 
n,  14).  The  maximum  ordinates  of  these  curves  are  given  by 
Eqs.  (130)  and  (132). 

For  a  uniform  load  of  p  pounds  per  foot,  extending  a  distance 
kl  from  either  end  of  the  main  span,  we  find,  by  integrating  the 
function  B(k)  in  Eq.  (129'), 


where  the  function, 

F(k)=%k2-%k*+k*,    ....     (133') 

and  may  be  obtained  directly  from  Table  I  or  from  the  graph  in 
Fig.  12.     For  k  =  i,  F(k)  =  i. 

For  similar  conditions  in  either  side  span,  we  find  for  a  loaded 
length  kih, 

-#i/,    -    -    (134) 


where  F(ki)  is  the  same  function  as  defined  by  Eq.  (133')  . 

The  horizontal  component  of  the  cable  tension  will  be  a 
maximum  when  all  spans  are  fully  loaded,  or  when  k  =  i  and 
ki  =  i.  Hence,  by  Eqs.  (133)  and  (134), 

Total  H  =  -~(i  +  2ir*v)pl.     ...     v    (135) 

For  a  live  load  covering  the  central  portion,  JK,  of  the  main 
span,  from  any  section  x=jl  to  any  other  section  x~kly  the 
application  of  Eq.  (133)  yields, 

l,    .     :..  (136) 


where  F(j)  and  F(k)  are  the  same  function  as  defined  by  Eq. 


TWO-HINGED   STIFFENING  TRUSSES  41 

The  graph  of  F(k)  in  Fig.  1  2  shows  the  proportional  increase 
in  the  value  of  H  as  a  uniform  load  comes  on  and  fills  the  main 
span  (or  either  side  span).  The  difference  between  the  two 

ordinates  for  any  sections,  /  and  K,  multiplied  by  ——for  by 

CL—  V  will  give  the  value  of  H  for  the  corresponding  partial 

loading  JK. 

For  opposite  loading  conditions,  that  is,  load  covering  both 
side  spans  and  all  of  the  main  span  with  the  exception  of  the 
central  portion  JK,  we  find  the  value  of  H  by  subtracting  the 
members  of  Eq.  (136)  from  those  of  Eq.  (135): 


20.  Moments  in  the  Stiffening  Truss.  —  The  bending  moment 
at  any  section  (main  or  side  span)  is  given  by  Eq.  (82), 

M  =  M'-Hy,    Mi=Mif-Hyi.    .     .      .     (138) 

If  any  span  is  free  from  load,  the  moments  for  that  span  are 
obtained  by  placing  M'  (or  Mi)  equal  to  zero,  giving, 

M=-Hy,      Mi=-Hyi,    ....     (139) 

where  H  is  the  cable  tension  produced  by  loads  in  the  other 
spans,  or  by  temperature. 

With  all  three  spans  loaded,  using  the  value  of  H  given  by 
Eq.  (135),  Eq.  (138)  yields,  for  any  section  in  the  main  span, 


Total  M  =  $px(l-x)i-(i  +  2if*v),    .      .     (140) 
and,  for  any  section  in  the  side  span, 

Total  Mi=%pxi(h-xi)\i-(i  +  2&>v)        .     (141) 


The  influence  diagrams  for  bending  moment  are  constructed, 
in  accordance  with  Eq.   (86),  by  superimposing  the  influence 

Mf 
triangle  for  —  on  the   ^-influence   curve:     The   H-curve  is 


42 


STRESSES   IN   SUSPENSION  BRIDGES 


TABLE  I 

FUNCTIONS  OCCURRING  IN  SUSPENSION  BRIDGE  FORMULAS 


H 

H  for 

Critical 

Influence 

Minimum  Moments 

Uniform 

Shears 

Points 

Line 

Loads 

k 

B(k) 

C(k) 

D(k] 

F(k) 

G(k) 

k 

k(l-2k*+k3) 

k+tf-k* 

(2-*-4*H-3*»)(i-*)* 

#2-5*4+46 

§(l-/fe)3-(r-jfe)2+I 

0 

O 

o 

2.0 

O 

0.4 

0 

•05 

.0498 

.0524 

I.75II 

.OO62 

.4404 

•05 

.1 

.0981 

.  1090 

I  .  5090 

.0248 

.4816 

.10 

•  15 

•1438 

.  1691 

1.2790 

•0550 

•5232 

•  15 

.2 

.1856 

.2320 

I  .  0650 

.0963 

.5648 

.  20 

•25 

.2227 

.2969 

.8704 

•1474 

.6062 

•  25 

•3 

.2541 

•  3630 

.6962 

.2O72 

.6472 

•  30 

-35 

•2793 

.4296 

•5445 

.2740 

.6874 

•35 

•  4 

.2976 

.4960 

•4147 

.3462 

.7264 

.40 

•45 

.3088 

•5614 

•3065 

.4222 

.7640 

•45 

•5 

•3125 

.6250 

.2188 

o-5 

.8000 

•So 

•55 

.3088 

.6861 

•1497 

•5778 

.8340 

•55 

.6 

.2976 

•7440 

•0973 

•  6538 

.8656 

.60 

•65 

•2793 

•7979 

•0593 

.7260 

.8946 

•65 

•  7 

•2541 

.8470 

•0332  . 

.7928 

.9208 

.70 

•75 

.2227 

.8906 

.0166 

•  8526 

.9438 

•75 

.8 

.1856 

.9280 

.0070 

•9037 

.9632 

.80 

•85 

•1438 

•9584 

.0023 

•  945° 

.9788 

•85 

•9 

.0981 

.9810 

.0005 

•9752 

.9904 

.90 

•95 

.0498 

•9951 

.0003 

•9938 

.9976 

•95 

i  .00 

O 

I.O 

o 

i  .0 

I  .0 

i  .0 

plotted  with  ordinates  given  by  Eqs.  (129)  and  (131);  the  — 

triangles  have  a  constant   height,  —  in  the    main    span  and 

47 

—    in  the  side   spans.     The   resulting   influence  diagrams  are 

4/i 

shown  in  Figs,  nb  and  nc.    The  intercepted  areas,  multiplied 

by  py,  give  the  desired  bending  moments;    areas  below  the 

F-curve  represent  positive  or  maximum  moments,  and  those 

above  represent  negative  or  minimum  moments. 


TWO-HINGED  STIFFENING  TRUSSES 


43 


LOO 


.90 


.30        .35        .40         .45         .50        .55        .60        .65        .70        .75        .80        .85         .90        .95        LOO 


C(k)=k+k2-k3- 


FIG.  13. — Graphs  for  the  Solution  of  Suspension  Bridge  Formulas. 

(Supplementary  to  Fig.  12). 


44  STRESSES  IN  SUSPENSION  BRIDGES 

For  any  section  in  the  main  span,  there  is  a  zero  point  or 
critical  point  K  (Fig.  116),  represented  by  the  intersection  of  the 
superimposed  influence  lines.  The  distance  kl  to  this  critical 
point  is  given  by  the  relation, 

C(k)=k+k2-k*  =  N-n.~.     .  ,     .     .     (142) 

Values  of  the  function  C(k)  are  listed  in  Table  i  and  plotted  in  a 
graph  in  Figs.  12  and  13,  to  facilitate  the  solution  of  Eq.  (142) 
for  k. 

^/The  maximum  negative  moment  at  any  section  of  the  main 
span  is  obtained  by  loading  the  length  l  —  kl  in  that  span  and 
completely  loading  both  side  spans  (Fig.  nb).  Then,  using  the 
values  of  Eqs.  (133)  and  (135),  Eq.  (138)  yields, 


,      Min.  Jf  =  -)+4*H     .     .     (143) 

J/-. 

where  the  function, 

A 

D(k)  =  (2-k-4k2+3k*)(i-k)2,     .      .     (1430 

and  is  given,  for  different  values  of  k,  by  Table  I  and  by  the 
graph  in  Fig.  12  or  13.  The  value  of  k  or  C(k)  obtained  from 
Eq.  (142)  is  to  be  used. 

Equation  (143)  applies  to  all  sections  from  x  =  o  to  x'  =  —  -I. 

4 

For  the  minimum  moments  at  the  sections  near  the  center,  from 
x'  to  l—x'j  it  is  necessary  to  bring  on  some  load  also  from  the  left 
end  of  the  span,  as  there  are  two  critical  points,  K'  and  Kn  ',  for 
these  sections  (see  dotted  diagram,  Fig.  n&);  so  that  the 
expression  (143)  for  these  moments  must  be  corrected  by  replac- 
ing D(k)  by  D(V)+D(k"),  where  k'  is  the  value  of  k  (Eq.  142) 
corresponding  to  the  given  section  x,  and  k"  is  the  value  of  k 
corresponding  to  the  symmetrically  located  section  (/—#). 

The  maximum  positive  moments  are  given  by  the  relation, 

Max.  M  =  Total  Jlf-Min.  M.     .      .     •  (144) 


TWO-HINGED  STIFFENING  TRUSSES  45 

Subtracting  the  values  given  by  Eq.  (143)  from  those  given  by 
Eq.  (140),  we  obtain, 

.     (144') 

The  loading  corresponding  to  this  moment  is  indicated  in  Fig. 
116;  only  a  portion  of  the  main  span  is  loaded,  the  side  spans 
being  without  load. 

There  are  no  critical  points  in  the  side  spans.  For  the  great- 
est negative  moment  at  any  section  x\  in  one  of  the  side  spans, 
load  the  other  two  spans  (Fig.  nc),  giving, 

(145) 

Loading  the  span  itself  produces  the  greatest  positive 
moments,  which  are  obtained  by  the  relation, 

Max.  Mi  =  Total  Mi  -  Min.  M  i  .....     (146) 

Subtracting  the  values  given  by  Eq.  (145)  from  those  given  by 
Eq.  (141),  we  obtain, 


The  maximum  and  minimum  moments  for  the  various 
sections  of  a  stiffening  truss  (Type  25),  as  calculated  from 
Eqs.  (143),  (144)?  (J45)  and  (146),  are  plotted  in  Fig.  nd,  to 
serve  as  a  guide  in  proportioning  the  chord  members. 

21.  Shears  in  the  Stiffening  Truss.  —  With  the  three  spans 
completely  loaded,  the  shear  at  any  section  x  in  the  main  span 
will  be,  by  Eqs.  (84),  (90)  and  (135), 


Total  V  =  ^p(l-2x)i-~(i  +  2ir3v)J      .      .     (147) 
and,  in  the  side  spans, 

Total  Fi-  iX/i-2*i)i----(i  +  2*«r)     .     (148) 


The  influence  diagram  for  shear  at  any  section  is  constructed 
according  to  Eq.  (89),  by  superimposing  on  che  #-curve  (Eqs. 


46 


STRESSES   IN   SUSPENSION  BRIDGES 


129  and  131)  the  influence  lines  for 


tan 


The  latter  will  have 


end  intercepts  =  cot  0,  where  <j>  is  the  slope  of  the  cable  at  the 
given  section.  The  resulting  influence  diagram  is  shown  in 
Fig.  140.  The  intercepted  areas,  multiplied  by  p  •  tan  $,  give 
the  desired  vertical  shears  V.  Areas  below  the  /7-curve  repre- 


A^ 
//ox 


FIG.  14. — Shear  Diagrams  for  Two-hinged  Stiffening  Truss. 
(Type  25). 

sent  positive  or  maximum  shears,  and  areas  above  represent 
negative  or  minimum  shears. 

Loading  the  main  span  from  the  given  section  X  to  the  end 
of  the  span,  we  obtain  the  maximum  positive  shears  by  Eqs. 
(84),  (90)  and  (133), 


TWO-HINGED   STIFFENING  TRUSSES  47 

where  the  function, 

2  f       ~ 

+I'-  •  •  •  (I49) 

and  is  given  by  Table  I  and  the  graph  in  Fig.  12. 

For  the  sections  near  the  ends  of  the  span,  from  x  =  o  to 

x  =  -\i  --  ),  the  loads  must  not  extend  to  the  end  of  the  span 

A       4/ 

to  produce  the  maximum  positive  shears,  but  must  extend  only 
to  a  point  K  (Fig.  140)  whose  abscissa  x  —  kl  is  determined  by 
the  following  equation: 

.      .      .     (I50) 


4    /  —  2# 

For  these  sections,  the  positive  shears  given  by  Eq.  (149)  must 
be  increased  by  an  amount, 


•-  Add.  F  =  ^/(i-*)2--~-G(*)-i,    -     (151) 

where  the  function, 

....     (151') 


and,  like  the  same  function  in  Eq.  (149'),  is  given  by  Table  I 
and  the  graph  in  Fig.  12  or  13. 

Formula  (150)  for  the  critical  section  is  solved  in  the  same 
manner  as  Eq.  (142)  with  the  aid  of  Table  I  or  the  graph  in 
Fig.  12  or  13. 

There  are  no  critical  points  for  shear  in  the  side  spans.  The 
influence  diagram  (Fig.  146)  shows  the  conditions  of  loading. 
For  maximum  shear  at  any  section  xi,  the  load  extends  from  the 
section  to  the  tower,  giving, 


where  G  (    -    is  the  same  function  as  denned  by  Eqs.  (1490  and 


48  STRESSES  IN   SUSPENSION  BRIDGES 

The  maximum  negative  shears  in  main  and  side  spans  are 
given  by  the  relations, 

Min.  V  =  Total  V  -Max.  F,    .     .     .     (153) 
and 

Min.  Fi=  Total  Fi-Max.  Fi.        .     .    (153') 

The  maximum  positive  and  negative  shears  for  different 
sections  of  the  main  and  side  spans,  as  given  by  Eqs.  (149), 
(152),  (153)  and  (1530?  are  plotted  for  a  typical  suspension 
bridge,  in  Fig.  140,  to  serve  as  a  guide  in  proportioning  the  web 
members. 

22.  Temperature  Stresses.  —  The  total  length  of  cable 
between  anchorages  is,  by  Eqs.  (22)  and  (26), 


.     (154) 
V  3  sec  ai/ 

Corrections  should  be  made  in  the  value  of  L  for  any  portions  of 
the  cable  not  included  in  the  spans  /  or  l\. 

Under  the  influence  of  a  rise  in  temperature,  the  total  increase 
in  length  between  anchorages  will  be: 

A  =  co/L  ........     (155) 

Substituting  this  value  for  the  numerator  in  Eqs.  (116)  to  (125), 
we  obtain, 

„         sEI-vtL  ,      . 

t==    ~  f2-N'l  '   -      '     '     '     '     '  S  ' 

where  N  denotes  the  denominator  of  Eq.  (125)  and  L  is  given 
by  Eq.  (154).  (For  an  extreme  variation  of  /=±6o°F., 
Eco/  =  11,  7  20.) 

The  resulting  bending  moment  at  any  section  of  the  truss  is 
given  by, 

Mt=-Ht.y,  .    V    .     ,'•  j&  .     (157) 

and  the  vertical  shear  by, 

F,=  -tf,(tan0-tana),       .      .     (158) 

where  0  is  the  inclination  of  the  cable  at  the  given  section,  and 
a  is  the  inclination  of  the  cable  chord  (Eqs.  84',  90). 

23.  Deflections  of  the  Stiffening  Truss.  —  For  any  specified 
loading,  the  deflections  of  the  stiffening  truss  may  be  computed 


TWO-HINGED  STIFFENING  TRUSSES  49 

as  the  difference  between  the  downward  deflections  produced  by 
the  applied  loads  and  the  upward  deflections  produced  by  the 
suspender  forces,  the  stiffening  truss  being  treated  as  a  simple 
beam  (for  Types  2F  and  25).  The  suspender  forces  are  equiva- 
lent to  an  upward-acting  load,  uniformly  distributed  over  the 
entire  span,  and,  by  Eq.  (78),  amounting  to, 

.     .  .    -   ,.'.  s=%.H.      ...     .     .     .     (78) 

For  a  uniform  load  p  covering  the  main  span,  the  resultant 
effective  load  acting  on  the  stiffening  truss  will  be,  by  Eqs.  (78) 
and  (135), 

......    (I59) 


and  the  resulting  deflection  will  be, 


In  the  general  case,  the  applied  loads  will  produce  a  deflection 
at  a  distance  x  of, 


The  suspender  forces,  given  by  Eq.  (78),  will  produce  an  upward 
deflection,  at  a  distance  x,  of, 

-H.     .     .     .     (162) 


It  should  be  noted  that  this  deflection  curve  (Eq.  162)  is  similar 
to  the  //-influence  curve  given  by  Eq.  (129).  Using  the  function 
defined  by  Eq.  (129'),  Eq.  (162)  may  be  written, 


The  resulting  deflection  of  the  truss  at  any  point  will  then  be 
obtained  from  Eqs.  (161)  and  (162)  as 

d  =  d'-d".        .     ,'    i  :.     .     .     (163) 


50 


STRESSES  IN  SUSPENSION  BRIDGES 


Equation  (160),  for  a  full-span  load,  may  be  derived  directly 
from  Eq.  (163). 

If  merely  the  half  -span  is  loaded  with  p  per  unit  length,  then 
the  deflection  at  the  quarter-point  will  be,  by  Eqs.  (161)  and 
(162),  in  the  loaded  half, 


and,  in  the  unloaded  half, 


/S7..L-  26 

' 


6i44\  2 


EI' 


(164') 


FIG.  15. — Detroit- Windsor  Bridge. 
(Type  2F). 

Span  1803  feet.  Combined  Railway  and  Highway  Bridge.  8  Cables.  C.  E.  Fowler, 
Chief  Engineer.  D.  B.  Steinman,  Associate  Engineer.  W.  H.  Burr,  G.  H.  Pegram, 
C.  N.  Monsarrat  and  C.  R.  Young,  Consulting  Engineers. 

By  Eq.  (125),  N  will  always  be  greater  than  f.  Substituting  this 
minimum  value  in  Eq.  (164)  or  (164'),  we  obtain  the  upward  or 
downward  deflections  at  the  quarter-points: 


TWO-HINGED  STIFFENING  TRUSSES 


51 


The  deflections  produced  by  temperature  effects,  or  by  a 
yielding  of  the  anchorages,  are  given  by  Eq.  (162'),  upon  sub- 
stituting for  H  the  horizontal  tension  caused  by  the  given 
influence.  Substituting  the  expression  from  Eq.  (156) ,  we  obtain, 


AZ, 

I]  N-n'  ' 


.     .     (166) 


where  the  function  B( - \  is  defined  by  Eq.  (129')  and  is  given  by 

Table  I  and  the  graph  in  Fig.  12. 

24.  Straight  Backstays  (Type  2F). — If  the  stiffening  truss  is 
built  independent  of  the  cables  in  the  side  spans  (Figs.  15,  16), 


T 


I 


I—  4—  H 

FIG.  1  6.  —  Two-hinged  Stiffening  Truss  with  Straight  Backstays. 
(Type  2F). 

the  backstays  will  be  straight  and  /i  =  o.    Consequently  all  terms 

containing  /i,  y\,  n\=^-,  or  v  =  ^  will  vanish  in  Eqs.  (125)  to 

*i  7 

(166)  inclusive. 

The  side  spans  will  then  act  as  simple  beams,  unaffected  by 
any  loads  in  the  other  spans;  and  the  main-span  and  cable 
stresses  will  be  unaffected  by  any  loads  in  the  side  spans. 

The  denominator  of  the  general  expression  for  H  (Eq.  125) 
will  then  reduce  to  the  denominator  of  Eq.  (127): 


Equations  (131),  (134),  and  (145),  will  vanish. 


*  See  Footnote  to  Eq.  127. 


52  STRESSES  IN  SUSPENSION  BRIDGES 

The  maximum  value  of  H  will  be  produced  by  a  uniform 
load  p  covering  the  main  span,  and  will  be,  by  Eq.  (135), 

Total 


. 

$N-n 

The  bending  moment  at  any  section  x  of  the  main  span  will 
then  be,  by  Eq.  (140), 

Total  M=±px(l-x)(i~\       .     .;  .     (169) 
The  greatest  negative  bending  moment  will  be,  by  Eq.  (143), 


-\/r*        ir  —          ^/LN  /        \ 

Mm.  Af=--^—  ^—  —  --D(k).       .,.;..    .     (170) 

The  greatest  positive  moment  is  then  given  by  Eq.  (144')  , 
or  by, 

Max.  M  =  Total  K-Min.  M.       ,     ^    .     (171) 

In  the  side  spans,  there  will  be  no  negative  moments.     The 
greatest  positive  moments  will  be,  by  Eq.  (141), 

Max.  Mi  =  Tota]Mi=%piXi(h-xi)  ,  .      .     (172) 

exactly  as  in  a  simple  beam. 

With  load  covering  the  entire  span,  the  shears  in  the  main 
span  will  be,  by  Eq.  (147), 

Total  7  =  }X*-2*)i~,  • 


and,  in  the  side  spans,  by  Eq.  (148), 

TotalFi=|#i(/i-2*i).      .     .     .  ;.     .     (174) 

The  maximum  shears  in  the  main  span  will  be  given  by  Eqs. 
(149),  (150)  and  (151).  In  the  side  spans,  the  maximum  shears 
will  be,  by  Eq.  (152), 

Max.  7i-&i/,(i-£l)2.   .     .     .     ,     .     (175) 

exactly  as  in  a  simple  beam. 

The  total  length  of  cable  will  be,  by  Eq.  (154), 

Z,  =  /'(i+fw2)  +  2/2-secai,       ,     .     (176) 

and  the  temperature  stresses  are  then  given  by  Eqs.  (156),  (157) 
and  (158). 


HINGELESS  STIFFENING  TRUSSES  53 

SECTION  VI.— HINGELESS  STIFFENING  TRUSSES 
(Types  QF  and  OS) 

25.  Fundamental  Relations. — Hingeless  stiffening  trusses 
(Figs.  17,  1 8)  are  continuous  at  the  towers;  hence  there  will 
be  bending  moments  in  the  truss  at  the  towers. 

The  moments  and  shears  at  any  section  in  the  stiffening  truss 
will  be  the  resultants  of  the  values  produced  by  the  downward- 
acting  loads  (Mr  and  V')  and  the  upward-acting  suspender 


FIG.  17. — Suspension  Bridge 'over  the  Rhine  at  Cologne. 
(Type  OS). 

Continuous  Stiffening  Girder.     Eyebar  Chains.      Self-anchored.     Rocker  Towers. 
Span  605  feet.    Erected  1915. 


forces  (Ms  and  Vs).  Equations  (78),  (79)  and  (80)  will  apply; 
but  the  continuity  of  the  truss  must  be  taken  into  account  in 
calculating  the  respective  moments  and  shears. 

If  MI  and  M2  are  the  bending  moments  at  the  towers  pro- 
duced by  the  downward  loads  on  the  stiffening  truss,  and  if  MQ 
is  the  simple-beam  bending  moment  at  any  section  x,  then  the 


54  STRESSES  IN  SUSPENSION  BRIDGES 

resultant  bending  moment  due  to  the  downward  loads  acting  on 
the  continuous  truss  will  be, 


il*       .         .         .        (I77) 

in  the  main  span,  and, 

M'  =  M0+^-Mi,2,      .....      (177') 

in  the  side  spans. 

The  upward-acting  suspender  forces  will  be  uniform  over 
each  span.     For  any  value  of  H,  by  Eq.  (78),  the  upward  pull 

8/"  8/" 

will  be  B~  per  lineal  foot  in  the  main  span  and  H--^  in  the 

side  spans.  Then,  by  the  Theorem  of  Three  Moments  for  uni- 
form load  conditions,  we  find  the  moments  at  the  towers  (for 
symmetrical  spans)  to  be, 

-H-mi  =  -H-m2=-H'(e-f),   .      .     .     (178) 
in  which  the  coefficient  of  /  is  a  constant  defined  by, 


where  i,  r,  and  v  are  defined  by  JEq.  (126). 

The  simple-beam  bending  m6nieiit  produced  by  the  suspender 
forces  is  given  by  Eq.  (81)  as  H-y.  Adding  the  correction  for 
the  end  moments  at  the  towers  (Eq.  178),  we  obtain  the  result- 
ant suspender  moments  as, 

M,=H-(y-e-f),      .....     (180) 

for  any  section  in  the  main  span  ;  and,  for  any  section  in  the  side- 
spans, 

(181) 


where  x\  is  measured  from  the  free  end  of  the  span,  and  yi  is  the 
vertical  ordinate  of  the  side  cable  below  the  connecting  chord 
D'A'  (Fig.  i8a). 


HINGELESS  STIFFENING  TRUSSES  55 

Substituting  (177),  (i77')>  (180)  and  (181)  in  Eq.  (79),  we 
have,  for  bending  moments  in  the  main  span,  (Fig.  19), 


.M2-H(y-ef),    .     .     (182) 
and,  for  bending  moments  in  the  side  span, 


, 

k  \         k 

If  any  span  is  without  load,  Mo  for  that  span  will  vanish. 
The  shears  produced  by  the  downward-acting  loads  will  be, 


=!     .....     (184) 
in  the  main  span,  and 

or     V'  =  V0-**,       .      .     (185) 


in  the  side  spans.     In  these  equations,  Fo  denotes  the  simple- 
beam  shears  for  the  given  loading. 

The  shears  produced  by  the  upward-acting  suspender  forces 
will  be 

Fs  =  #(tan</>-tana)       .....     (186) 

in  the  main  span,  and 

/  rf\ 

VS  =  H  tan</>i-tan«i-fM       .      .      .     (187) 

\  hi 

in  the  side  spans. 

Substituting  (184),  (185),  (186)  and  (187)  in  Eq.  (80),  we 
have,  for  resultant  shears  in  the  main  span, 

.     .     (188) 


and,  for  resultant  shears  in  the  side  spans, 


(189) 
k/ 

If  any  span  is  without  load,  Fo  for  that  span  will  vanish. 
If  the  two  towers  are  of  equal  height,  then,  in  the  main  span, 


56  STRESSES  IN  SUSPENSION  BRIDGES 

26.  Moments  at  the  Towers  (Types  OF  and  OS).—  The 
values  of  the  end-moments  M\  and  M.%,  used  in  Eqs.  (177)  to 
(189),  may  be  determined,  for  any  given  loading,  by  the  Theorem 
of  Three  Moments. 

For  a  concentration  P  in  the  main  span,  at  a  distance  kl  from 
the  left  tower,  we  thus  obtain, 


T/       7\    i--         --  /      N 

M2=-Pl-k(i-k)  ^6      .    ;  --  —.  .      .     (191) 
(3  +  2w)(i  +  2*r) 

The  sum  of  these  two  end-moments  will  be, 

-k)  f             x 

.r      ,      .      ....     (192) 

.  .      .     (193) 


_ 
and  the  difference  will  be, 


For  a  concentration  P  in  the  left  side  span,  at  a  distance  kl\ 
from  the  outer  end,  the  Theorem  of  Three  Moments  yields: 

.  .  .  (I94) 


ir2(k-k3) 


For  a  uniform  load  covering  the  main  span,  we  obtain, 

Ml=M2=-    ,P^.Y        .....     (196) 
4(3  -Mr) 

For  a  uniform  load  covering  the  left  side  span,  we  obtain, 


.     ,     .     (I98) 


4 

For  a  uniform  load  covering  all  three  spans,  we  obtain, 

•     .     .     .     (I99) 


HINGELESS  STIFFENING  TRUSSES  57 

27.  The  Horizontal  Tension  H.  —  The  general  formula 
(117)  for  the  horizontal  tension  H  is  applicable  to  the  continu- 
ous stiffening  truss  (Types  OF  and  05). 

Equation  (118),  for  bending  moments  produced  by  the 
suspender  forces,  is  now  replaced  by  the  expressions  (180)  and 
(181),  and  Eq.  (119)  becomes, 

m=-y+ef,  ......     (200) 

for  the  main  span,  and 

-ef,  .....     (201) 


for  the  side  spans. 

Substituting  these  values  and  integrating  over  all  three  spans, 
we  obtain,  as  a  substitute  for  Eq.  (122), 


(202) 


Equations  (120),  (123),  (124)  and  (1240  are  retained 
unchanged.  Collecting  all  the  values  and  substituting  in  Eq. 
(117),  we  obtain  the  expression  for  H  in  the  continuous  type  of 
suspension  bridge  (in  place  of  Eq.  125)  : 


H  =  J  H>x° — i li-i A-  (201) 

-   '    -9   '    --•"/8-9   '    ^-2CT) 


The  denominator  of  this  expression  is  a  constant  for  a  given 
structure,  and  will  henceforth  be  denoted  by  N.  (If  hinges  are 
inserted  at  the  towers,  the  coefficient  of  continuity  e  will  be 
zero,  and  Eq.  [203]  reduces  to  Eq.  [125]). 

28.  Values  of  H  for  Special  Cases  of  Loading. — In  the  last 
equation  (203),  the  value  of  the  numerator  depends  upon  the 
loading  in  any  particular  case.  Expressing  M'  as  a  function  of  x 
(Eq.  177),  substituting  the  value  of  y  .given  by  Eq.  (14),  and 

*  See  Footnote  to  Eq.  (125).  » 


58 


STRESSES  IN   SUSPENSION  BRIDGES 


performing  the  integration  as  indicated,  we  find,  for  a  single 
load  P  at  a  distance  kl  from  either  end  of  the  main  span, 


,      .      .      .     (204) 


where  N  denotes  the  denominator  of  Eq.  (203),  and  the  function 
B(k)  is  defined  by  Eq.  (129')  and  is  given  by  Table  I  and  Fig.  12. 


FIG.  1  8.  —  Moment  Diagram  for  Continuous  Stiffening  Truss. 
(Type  OS). 

Similarly,  for  a  concentration  PI  in  either  side  span,  at  a 
distance  k\l\  from  the  free  end,  we  obtain, 


.      (205) 


Plotting  Eqs.  (204)  and  (205),  we  obtain  the  ^-influence 
line,  Fig.  186. 

If  the  main  span  is  completely  loaded,  we  obtain,  by  inte- 
grating Eq.  (204), 

H  =  ^-(---}'pl.  (206) 


HINGELESS  STIFFENING  TRUSSES  59 

If  both  side  spans  are  completely  loaded,  we  obtain,  by 
integrating  Eq.  (205), 


If  the  main  span  is  loaded  for  a  distance  kl  from  either  tower, 
we  obtain,  from  Eq.  (204), 

,     .      .     (208) 


where  F(k)  is  defined  by  Eq.  (133')  and  is  given  by  Table  I  and 
Fig.  12. 

If  either  side  span  is  loaded  for  a  distance  k\h  from  the  free 
end,  we  obtain,  from  Eq.  (205), 

-ie(2-kl2)'kl2]p1l,    .     (209) 


where  F(ki)  is  the  same  function  as  defined  by  Eq.  (1330. 

In  the  foregoing  equations,  N  represents  the  denominator  of 
Eq.  (203). 

(If  the  stiffening  truss  is  interrupted  at  the  towers,  the  factor 
of  continuity  e  =  o,  and  the  above  formulas  reduce  to  the  cor- 
responding equations  [129]  to  [135]  for  the  two-hinged  stiffening 
truss.) 

29.  Moments  in  the  Stiffening  Truss.  —  With  all  three  spans 
loaded,  the  bending  moment  at  any  section  of  the  main  span  is 
given,  very  closely,  by  Eqs.  (182)  and  (199),  as, 

Total  M=±p-H-4x-(l-x)-e(±pP-H.f),      (210) 


and,  at  any  section  of  the  side  span  distant  x\  from  the  free  end 
by  Eqs.  (183)  and  (199),  as, 

Total  M=\p-Hxl(ll-xl}-e(lpP-H})         (211) 


where  e  is  defined  by  Eq.  (179),  and  H  is  given  by  the  combina- 
tion of  Eqs.  (206)  and  (207). 

The  moments  for  other  loadings  must  be  calculated  by  the 


60  STRESSES   IN  SUSPENSION  BRIDGES 

general  Eqs.  (182)  and  (183),  with  the  values  of  H  given  by  Eqs. 
(204)  to  (209),  and  the  values  of  M\  and  M2  given  by  Eqs.  (190) 
to  (199). 

Influence  lines  for  moments  may  be  drawn  as  in  the  previous 
cases.  For  moments  in  the  main  span,  Eq.  (182)  is  written  in 
the  form, 


M=  y-ef  -H-(y-ef),      .     (212) 

thus  giving  the  bending  moments  as  (y  —  ef)  times  the  intercepts 

1*7 

obtained  by  superimposing  the  influence  line  for  -  —  -  upon  the 

y-ef 

influence  line  for  H.  This  construction  is  indicated  in  Fig.  i8&. 
For  moments  in  the  side  spans,  the  corresponding  influence  line 
equation  is  obtained  from  Eq.  (183): 


M  = 


x\ 


H 


(213) 


For  the  continuous  stiffening  truss,  the  influence  line  method 
just  outlined  is  not  very  convenient,  as  the  M'  influence  line 
(Fig.  1 86)  is  a  curve  for  which  there  is  no  simple,  direct  method 
of  plotting. 

A  more  convenient  method  is  that  of  the  Equilibrium  Polygon 
constructed  with  pole-distance  H,  corresponding  to  Eq.  (85) 
and  Fig.  7.  For  the  continuous  stiffening  truss,  this  construc- 
tion is  modified  as  follows  (Fig.  19):  At  a  distance  cf  below  the 
closing  chord  A'B' ',  a  base  line  AB  is  drawn,  so  that  the  cable 
ordinates  measured  from  this  base  line  will  be  (y—cf)  and  will 
therefore  represent  Ms  (Eq.  180).  The  equilibrium  polygon 
A" MB"  for  any  given  loads  is  then  constructed  upon  the  same 
base  line,  with  the  same  pole-distance  77;  the  height  A  A" 
represents  —M\,  the  height  BB"  represents  —Mz,  and  the  poly- 
gon ordinates  below  A"B"  represent  M0:  hence,  by  Eq.  (177), 
the  ordinates  measured  below  the  base  line  AB  represent  M'. 
Then,  by  Eq.  (79),  the  intercept  between  the  cable  curve  and 


HINGELESS  STIFFENING  TRUSSES 


61 


the  superimposed  equilibrium  polygon,  multiplied  by  H,  will  give 
the  resultant  bending  moment  M  at  any  section. 

For  a  single  concentrated  load  P,  the  equilibrium  polygon 
A" MB"  is  a  triangle,  and  the  M  intercepts  can  easily  be  scaled 
or  figured.  By  moving  a  unit  load  P  to  successive  panel  points, 
we  thus  obtain  a  set  of  influence  values  of  M  for  all  sections. 

The  corresponding  construction  in  the  side  spans  is  also  indi- 
cated in  Fig.  19. 

30.  Temperature  Stresses. — The  horizontal  tension  pro- 
duced by  a  rise  in  temperature  of  f  is  given  by, 


Ht=- 


r 


f-N-r 


i_ 


FIG.  19. — Equilibrium  Polygon  for  Continuous  Stiffening  Truss. 
(Type  OS). 

where  N  is  the  denominator  of  Eq.  (203),  and  L  is  given  by 
Eq.  (154). 

The  resulting  moments  in  the  stiffening  truss  will  be  given, 
by  Eqs.  (180)  and  (181),  as, 

Mt=-Hr(y-ef), (215) 

for  the  main  span,  and, 

M^-H.fy.-^-.efJ,     .'    .      .     .     (216) 

for  the  side  spans. 

The  vertical  shears  are  given  by  Eqs.  (186)  and  (187)  as, 

7,=  -#«(tan0-tana),       ^     f     .      .     (217) 


62  STRESSES   IN   SUSPENSION   BRIDGES 

for  the  main  span,  and 

/  ef\ 

Ft=-ZMtan</>i-tanai-f-),    .      .      .     (218) 
\  hi 

for  the  side  spans. 

31.  Straight  Backstays  (Type  OF).— If  the  stiffening  truss 
in  the  side  spans  is  built  independent  of  the  cable  (Fig.  20),  the 
backstays  will  be  straight  and/i=o.  Consequently,  all  terms 

containing /i,  yi,  n\=^-,  or  i)  =  ^,  will  vanish  in  Eqs.  (177)  to 

k  J 

(218),  inclusive. 

On  account  of  the  continuity  of  the  trusses,  however,  each 
span  will  be  affected  by  loads  in  the  other  spans. 


i _ 


f 
f 


FIG.  20. — Continuous  Stiffening  Truss  with  Straight  Backstays. 
(Type  OF}. 

The  denominator  of  the  expression  for  £T,  Eq.  (203),  will 
become, 


where  e,  the  factor  of  continuity,  now  has  the  value, 

e  = r (220) 

Equation  (183),   for  bending  moments  in  the  side  spans, 
will  become, 

r'ef>      •     •     •     (221) 


See  Footnote  to  Eq.  (127). 


BRACED-CHAIN  SUSPENSION  BRIDGES  63 

and  Eq.  (189),  for  shears  in  the  side  spans,  will  become, 

.  .        .        .        -        (222) 


I  I 

For  a  concentration  PI  in  either  side  span,  Eq.  (205)  becomes, 

k^-Pl.       .     .     (223) 


For  a  uniform  load  covering  both  side  spans,  Eq.   (207) 
becomes, 


For  a  uniform  load  in  either  side  span,  covering  a  length 
from  the  free  end,  Eq.  (209)  becomes, 


For  a  uniform  load  covering  all  three  spans,  Eq.  (211),  for 
the  bending  moments  in  the  side  spans,  becomes 

Total  M  =  §pxi(h-xi)-^($pP-Hf).       .     (226) 

/i 

Equation  (216),  for  temperature  moments  in  the  side  spans, 
becomes, 

,    .....     (227) 


and  Eq.  (218),  for  the  shears,  becomes, 

....  (228) 


.    ..... 

h 

SECTION  VII.—  BRACED-CHAIN  SUSPENSION  BRIDGES 

32.  Three-hinged  Type  (SB).—  The  three-hinged  type  of 
braced-chain  suspension  bridge  is  statically  determinate.  The 
suspension  system  in  the  main  span  is  simply  an  inverted  three- 
hinged  arch.  The  equilibrium  polygon  for  any  applied  loading 
will  always  pass  through  the  three  hinges.  The  //-influence 
line  for  vertical  loads  reduces  to  a  triangle  whose  altitude,  if 


64 


STRESSES   IN   SUSPENSION  BRIDGES 


the  crown-hinge  is  at  the  middle  of  the  span  and  if  the  correspond- 
ing sag  is  denoted  by/,  is, 

#  =  -.'..,-..      .      .      .     (229) 

The  determination  of  the  stresses  is  made,  either  analytically 
or  graphically,  exactly  as  for  a  three-hinged  arch. 

Figure  21  shows  the  single-span  type,  in  which  the  backstays 
are  straight  (Type  3BF) .  If  the  lower  chord  is  made  to  coincide 
with  the  equilibrium  polygon  for  dead  load  or  full  live  load,  the 
stresses  in  the  top  chord  and  the  web  members  will  be  zero  for 
such  loading  conditions.  These  members  will  then  be  stressed 
only  by  partial  or  non-uniform  loading.  Under  partial  loading, 


FIG.  21. — Three-hinged  Braced  Chain  with  Straight  Backstays. 
(Type  3BF). 

the  equilibrium  polygon  will  be  displaced  from  coincidence  with 
the  lower  chord:  where  it  passes  between  the  two  chords,  both 
will  be  in  tension;  where  it  passes  below  the  bottom  chord, 
this  member  will  be  in  tension  and  the  top  chord  will  be  in  com- 
pression. If  the  curve  of  the  bottom  chord  is  made  such  that 
the  equilibrium  polygon  will  fall  near  the  center  of  the  truss  or 
between  the  two  chords  under  all  conditions  of  loading,  the 
stresses  in  both  chords  will  always  be  tension. 

Figure  22  shows  the  three-hinged  braced-chain  type  of  suspen- 
sion bridge  provided  with  side  spans  (Type  3BS).  The  stresses 
in  the  main  span  trusses  are  not  affected  by  the  presence  of  the 
side  spans,  and  are  found  as  outlined  above.  The  stresses  in 
the  side  spans  are  found  as  for  simple  truss  spans  of  the  same 
length,  excepting  that  there  must  be  added  the  stresses  due  to 


BRACED-CHAIN  SUSPENSION  BRIDGES 


65 


the  top  chord  acting  as  a  backstay  for  the  main  span.  This  top 
chord  receives  its  greatest  compression  when  the  span  in  ques- 
tion is  fully  loaded,  and  its  greatest  tension  when  the  main  span 
is  fully  loaded. 

Temperature  stresses  and  deflection  stresses  in  three-hinged 
structures  are  generally  neglected. 


FIG.  22. — Three-hinged  Braced  Chain  with  Side  Spans. 
(Type  3BS). 

33.  Two-hinged  Type  (25).— This  system  (Fig.  23)  is  static- 
ally of  single  indetermination  with  reference  to  the  external 
forces,  so  that  the  elastic  deformations  must  be  considered  in 
determining  the  unknown  reaction. 

The  structure  is  virtually  a  series  of  three   inverted  two- 


FIG.  23. — Two-hinged  Braced  Chain  with  Side  Spans 
(Type  2BS). 


H 


hinged  arch  trusses,  having  a  common  horizontal  tension 
resisted  by  the  anchorage. 

The  value  of  H  may  be  determined  by  the  same  method  as 
was  used  for  writing  Eqs.  (116)  and  (117).  In  this  case,  the 
general  equation  for  H  takes  the  form, 


66  STRESSES   IN   SUSPENSION  BRIDGES 

where  Z  denotes  the  stresses  in  the  members  for  any  external 
loading  when  H  =  o  (i.e.,  when  the  system  is  cut  at  the  anchor- 
ages) ;  u  denotes  the  stresses  produced  under  zero  loading  when 
H  =  i  ;  /  denotes  the  lengths  of  the  respective  members  and  A 
their  cross-sections.  The  summations  embrace  all  the  members 
in  the  entire  system  between  anchorages. 

The  stress  in  any  member  is  given  by  adding  to  Z  the  stress 
produced  by  H,  or, 

S  =  Z+H.u  ......     (231) 

Zl 

For  a  rise  in  temperature,  the  elastic  elongations  —  —  are  replaced 

iLA. 

by  thermal  elongations  w#,  and  Eq.  (230)  becomes, 
„     A 


For  uniform  temperature  rise  in  all  the  members,  Eq.  (232) 
may  be  written, 


where  L  is  the  total  horizontal  length  between  anchorages. 

Equations  (230)  to  (233)  may  also  be  used  for  the  ordinary 
types  of  suspension  bridge  with  straight  stiffening  truss  (Types 
2F  and  2S)  if  the  summations  are  applied  to  the  individual 
members  of  the  stiffening  truss  and  to  the  segments  of  the  cable 
between  hangers.  (The  hangers  and  towers  may  also  be 
included.)  This  will  give  more  accurate  results  than  the  ordi- 
nary method,  as  it  takes  into  account  the  varying  moments  of 
inertia  of  the  stiffening  truss  and  any  variations  from  parabolic 
form  of  cable. 

A  graphic  method  of  determining  H  is  to  find  the  vertical 
deflections  at  all  the  panel  points  produced  by  a  unit  horizontal 
force  (H  =  i)  applied  at  the  ends  of  the  system.  The  resulting 
deflection  curve  will  be  the  influence  line  for  H.  If  the  ordinates 
of  this  curve  are  divided  by  the  constant  5  (the  horizontal  dis- 
placement of  the  ends  of  the  system  produced  by  the  same  force 


BRACED-CHAIN  SUSPENSION  BRIDGES 


67 


H  =  i),  they  will  give  directly  the  values  of  H  produced  by  a 
unit  vertical  load  moving  over  the  spans. 

34.  Hingeless  Type  (OB). — This  type  of  suspension  bridge 
(Fig.  24)  is  threefold  statically  indeterminate,  the  redundant 
unknowns  being  the  horizontal  tension  H  and  the  moments  at 
the  towers.  Instead,  the  stresses  in  any  three  members,  such  as 
the  members  at  the  tops  of  the  towers  and  one  at  the  center  of 
the  main  span,  may  be  chosen  as  redundants.  Let  the  stresses 
in  the  three  redundant  members  under  any  given  loading  be 
denoted  by  Xi,  X2,  X^.  When  these  three  members  are  cut, 
the  structure  is  a  simple  three-hinged  arch;  in  this  condition, 
let  Z  denote  the  stresses  produced  by  the  external  loads,  and  let 
#1,  U2  and  u%  denote  the  stresses  produced  by  applying  internal 


FIG.  24. — Hingeless  Braced  Chain  Suspension  Bridge. 
(Type  OB  . 

forces    Xi  =  i,    X2  =  i,    and    X3  =  i.     Then,    when    the    three 
redundants  are  restored,  the  stress  in  any  member  will  be, 

S  =  Z+XiUi+X2U2+XzU3.       ,     .      .     (234) 

The  restoration  of  the  redundant  members  must  satisfy  the 
three  conditions, 


/      \ 
=O>     (235) 


^=o,     and 


EA 


which,  with  the  aid  of  Eq.  (234),  may  be  written: 


EA 


EA 


" 


=  o 


EA 


•  (236) 


UNIVERSITY  OF  CAL1FOHNIA 
n«ARTMENT  OF  CIVIC 


68  STRESSES   IN   SUSPENSION   BRIDGES 

The  redundant  members  are  to  be  included  in  these  summa- 
tions. 

The  solution  of  these  three  simultaneous  equations  will 
yield  the  three  unknowns  Xi,  X%  and  ^3,  and  their  sub- 
stitution in  Eq.  (234)  will  give  the  stresses  throughout  the 
structure. 


CHAPTER  II 
TYPES  AND  DETAILS  OF  CONSTRUCTION 

1.  Introduction. — The  economic  utilization  of  the  materials 
of  construction  demands  that,  as  far  as  possible,  the  predomi- 
nating stresses  in  any  structure  should  be  those  for  which  the 
material  is  best  adapted.  The  superior  economy  of  steel  in 
tension  and  the  uncertainties  involved  in  the  design  of  large- 
sized  compression  members  point  emphatically  to  the  conclusion 
that  the  material  of  long-span  bridges,  for  economic  designs, 
must  be  found  to  the  greatest  possible  extent  in  tensile  stress. 
This  requirement  is  best  fulfilled  by  the  suspension-bridge  type. 

The  superior  economy  of  the  suspension  type  for  long-span 
bridges  is  due  fundamentally  to  the  following  causes: 

1 .  The  very  direct  stress-paths  from  the  points  of  loading  to 
the  points  of  support. 

2.  The  predominance  of  tensile  stress. 

3.  The  highly  increased  ultimate  resistance  of  steel  in  the 
form  of  cable  wire. 

For  heavy  railway  bridges,  the  suspension  bridge  will  be 
more  economical  than  any  other  type  for  spans  exceeding  about 
1 500  feet.  As  the  live  load  becomes  lighter  in  proportion  to  the 
dead  load,  the  suspension  bridge  becomes  increasingly  economi- 
cal in  comparison  with  other  types.  For  light  highway  struc- 
tures, the  suspension  type  can  be  used  with  economic  justification 
for  spans  as  low  as  400  feet. 

Besides  the  economic  considerations,  the  suspension  bridge 
has  many  other  points  of  superiority.  It  is  light,  aesthetic, 
graceful;  it  provides  a  roadway  at  low  elevation,  and  it  has  a  low 
center  of  wind  pressure;  it  dispenses  with  falsework,  and  is 
easily  constructed,  using  materials  that  are  easily  transported; 

69 


70 


TYPES   OF   SUSPENSION  BRIDGES 


there  is  no  danger  of  failure  during  erection;    and  after  com- 
pletion, it  is  the  safest  structure  known  to  bridge  engineers. 

The  principal  carrying  member  is  the  cable,  and  this  has  a 
vast  reserve  of  strength.     In  other  structures,  the  failure  of  a 


single  truss  member  will  precipitate  a  collapse;  in  a  suspension 
bridge,  the  rest  of  the  structure  will  be  unaffected.  In  the  old 
Niagara  Railway  Suspension  Bridge  (built  1855),  the  chords  of 
the  stiffening  truss  were  broken  (due  to  overloading)  and  repaired 
repeatedly,  without  interrupting  the  railroad  traffic. 


CLASSIFICATION  TABLE 


71 


tsburgh 


bec  Design 


fc,  CO  ft,  O)  ft,  CO 

ft,  ft,  ^  ^  fcq  Scj 

cq  cq  cq  cq  cq  cq 

CN  55  <N  ?5  55  <N 


1 

CD 

fli 


18 
| 

fl 


en     en     en 


en     en     en     en     en 


§§ 


a 


tSentHen^en  C3en 


cucucucucucu  cucu 

CO   CO   CO   CO   CO   CO  CO   CO 


-o       « 


X 

rt 
> 


-g  CQ 
S   <N 

W 


72 


TYPES  OF  SUSPENSION  BRIDGES 


There  are  two  main  classes  of  suspension  bridges :  those  with 
suspended  stiffening  truss  (Figs.  25  to  36),  and  those  with  over- 
head braced-chain  construction  (Figs.  37  to  41).  For  purpose  of 
reference,  there  is  given  here  (page  71)  a  comprehensive  system 
of  classification  of  suspension  bridges,  with  mnemonic  type 
symbols  and  outstanding  examples. 

2.  Various  Arrangements  of  Suspension  Spans. — The  sim- 
plest form  of  suspension  bridge  is  a  single  span  (Type  2F  or  3F) 


FIG.  26. — Brooklyn  Bridge. 
(Type  3SD). 

Elevation,  Plan,  and  Cross-section. 

with  the  cable  carried  past  the  towers  as  diagonal  backstays 
(Figs.  27,  29).  If  side  spans  are  added  (Fig.  28),  they  are  inde- 
pendent of  the  cable  and  of  the  main  span.  The  single-span 
suspension  bridge  may  be  built  either  with  or  without  a  stiffen- 
ing truss  (Fig.  27). 

The  next  form  is  the  bridge  having  three  suspended  spans 
(Types  05,  IS,  2S,  35).  In  this  form,  stiffening  trusses  (or 
girders)  are  indispensable.  Only  two  towers  are  required,  and 
each  side  span  is  about  one-half  the  length  of  the  main  span 
(Figs.  10,  17,  25,30,33,35). 

If  the  main  span  is  provided  with  a  center  hinge  (in  addition 
to  end  hinges),  the  three-span  structure  becomes  statically 
determinate  (Type  35,  Fig.  26).  The  side  spans  are  suspended 


B 


74  TYPES  OF  SUSPENSION  BRIDGES 

from  the  cable,  but  carry  their  loads  as  simple  beams  without 
affecting  the  stresses  in  the  cable  or  in  the  main  span;  on  the 
other  hand,  any  load  in  the  main  span  or  tension  in  the  cable 
will  produce  relieving  stresses  in  the  side  spans. 

Multiple-span  suspension  construction,  with  more  than  two 
towers,  is  not  efficient;  the  great  economy  of  the  suspension 
type  is  lost.  As  the  number  of  spans  increases,  the  value  of  the 
cable  tension  H  is  proportionately  reduced,  and  more  of  the 
load  is  thrown  upon  the  stiffening  trusses ;  the  bending  moments 
and  the  deflections  are  thus  greatly  increased.  Examples  of 
this  type  are  the  Lambeth  Bridge,  London,  with  three  equal 
spans  of  280  feet;  and  the  Seventh  St.  Bridge,  Pittsburgh,  having 
two  main  spans  of  330  feet  and  two  side  spans  of  165  feet  each. 

Multiple-span  suspension  designs  have  been  proposed  with 
the  intermediate  piers  serving  as  anchorages  for  adjoining  spans. 
This  has  both  economic  and  aesthetic  disadvantages. 

A  suspension  bridge  of  two  spans  with  a  single  tower  would 
not  be  economical.  The  tower  would  have  to  be  twice  the 
normal  height  to  give  the  desired  sag-ratio  for  the  cables. 

3.  Wire  Cables  vs.  Eyebar  Chains. — One  of  the  first  ques- 
tions to  be  decided  in  the  design  of  a  suspension  bridge  is  the 
choice  between  a  wire  cable  and  a  chain  of  eyebars  (or  flats)  for 
the  principal  carrying  member.  The  latter  enables  the  bracing 
for  the  prevention  of  deformation  under  moving  load  to  be 
incorporated  in  the  suspension  system;  the  other  ordinarily 
requires  a  separate  stiffening  truss  for  the  reduction  of  these 
deflections. 

The  earliest  suspension  bridges  were  built  with  chains.  At 
first  (1796)  forged  wrought-iron  links  were  employed;  then  (1818) 
wrought-iron  eyebars  were  introduced;  and  later  (1828)  open- 
hearth  steel  eyebars  came  into  use.  John  A.  Roebling  estab- 
lished the  use  of  wire  cables  (about  1845);  and  since  his  time, 
wire  cables  have  been  used  in  practically  all  suspension  bridges. 
(Two  notable  exceptions  are  the  Elizabeth  Bridge  .at  Budapest 
[Fig.  34]  and  the  Rhine  Bridge  at  Cologne  [Fig.  17]). 

In  Lindenthal's  first  Quebec  Design  (Fig.  39),  and  in  his 
1894  design  for  the  projected  Hudson  River  Bridge,  he  pro- 


WIRE  CABLES   VS.  EYEBAR  CHAINS 


75 


posed  building  the  braced  chains 
of  pin-connected  wire  links;  such 
construction  would  have  the  ad- 
vantages of  accurate  work  and 
close  inspection  in  the  shop, 
rapid  erection,  and  possibility 
of  varying  the  cable-section  as 
required.  Thus  far  there  has 
been  no  opportunity,  however, 
of  demonstrating  the  feasibility 
of  combining  the  overhead  brac- 
ing system  with  a  cable  or  chain 
of  wire. 

The  economic  comparison  of 
wire  cable  and  eyebar  construc- 
tion rests  on  the  following  con- 
siderations: Steel  wire  with  an 
elastic  limit  of  150,000  pounds 
per  square  inch  costs  but  twice 
as  much  as  nickel-steel  eyebars 
with  one-third  the  elastic  limit. 
The  eyebar  heads  and  pins  add 
about  20  per  cent  to  the  weight 
of  the  chain.  The  wire  cable  is 
self-supporting  during  erection 
and  all  the  problems  involved 
have  been  worked  out  and  suc- 
cessfully demonstrated.  The 
eyebars,  on  the  other  hand,  unless 
expensive  falsework  is  used, 
would  require  temporary  sup- 
porting cables;  and  the  manu- 
facture and  erection  of  eyebars 
of  suitable  size  for  very  long 
spans  present  many  unsolved 
difficulties. 

The    following   points   have 


76  TYPES  OF  SUSPENSION  BRIDGES 

been  advanced  in  favor  of  chain  construction:  The  section  of 
an  eyebar  chain  may  be  varied  with  the  stress,  whereas  the  entire 
wire  cable  must  have  the  maximum  section.  The  possibility  of 
corrosion  in  wire  cables  (if  not  properly  protected)  and  of 
unequal  stressing  of  the  wires  (if  not  carefully  strung)  are 
further  arguments  for  adopting  eyebars.  Finally,  pin-connected 
eyebars  permit  speedier  erection,  especially  in  spans  which  would 
require  large-sized  cables. 

Comparative  designs  have  shown  that,  although  the  eyebar 
chain  is  2  to  i\  times  as  heavy  as  the  wire  cable,  the  difference 
in  cost  is  generally  very  small.  .  Where  two  designs  are  of  equal 
cost,  the  heavier  bridge  is  to  be  preferred  as  giving  a  more  rigid 
structure.  Greater  weight,  if  it  does  not  increase  the  cost,  is  an 
advantage  in  a  bridge,  as  it  serves  to  increase  the  rigidity  and 
the  life  of  the  structure. 

With  present  materials  and  prices,  chain  construction 
becomes  more  expensive  than  wire  cables  at  about  icoo-foot 
span.  (See  Fig.  28.)  The  development  of  high  alloy  steels 
at  a  sufficiently  low  unit  price  may,  however,  enable  eyebar 
construction  to  displace  wire  cables  even  in  the  longest 
spans. 

For  the  proposed  Hudson  River  Bridge  of  3240  feet  span 
(Frontispiece  and  Fig.  41),  it  was  found  that  the  adoption  of  the 
overhead  bracing  system  instead  of  a  suspended  stiffening  truss 
yielded  a  saving  which  greatly  outweighed  the  extra  cost  of 
employing  eyebars  instead  of  wire  cables. 

4.  Methods  of  Vertical  Stiffening. — On  account  of  the 
deformations  and  undulations  under  moving  load,  unstiffened 
suspension  bridges  should  not  be  used  except  for  footbridges. 

If  no  stiffening  truss  is  provided,  the  distortions  and  oscilla- 
tions of  the  cable  may  be  limited  by  using  a  small  sag-ratio; 
by  making  the  floor  deep  and  continuous;  or  by  employing  a 
latticed  railing  as  a  stiffening  construction  (Figs.  27,  33). 

Another  method  of  stiffening  the  suspension  bridge  is  by  the 
introduction  of  diagonal  stays  between  the  tower  and  the  road- 
way (Fig.  26).  These,  however,  have  the  disadvantage  of  making 
the  stress-action  uncertain,  and  of  becoming  either  overstressed 


METHODS  OF  VERTICAL  STIFFENING 


77 


or  inoperative  under  changes  of  temperature;  moreover,  they 
introduce  unbalanced  stresses  in  the  towers. 

In  recent  French  construction,  diagonal  stays  are  utilized, 
but  the  redundancy  of  members  is  more  or  less  remedied  by 
omitting  the  suspenders  near  the  towers  (Fig.  29).  The  inde- 
terminateness  is  thus  relieved,  and  the  cable  stress  is  reduced. 
This  arrangement  may  be  used  to  advantage  in  the  reconstruc- 
tion of  weak  suspension  bridges. 

A  different  method  of  vertical  stiffening,  known  as  the  Ordish- 
Lefeuvre  System,  dispenses  with  cable  and  suspenders;  it  con- 
sists of  diagonal  stays  running  from  the  tops  of  the  towers  and 
meeting  at  a  number  of  points  along  the  span,  so  as  to  provide  a 
triangular  suspension  for  each  point.  These  diagonal  stays  are 


FIG.  29. — Suspension  Bridge  at  Cannes-Ecluse. 
(Type  2FD). 

Over  the  Yonne  River  (France).    Span  760  feet.     Built   1900.  Wire  Rope  Cables. 

Diagonal  Stays. 

held  straight  by  hangers  from  a  light  catenary  cable  overhead. 
This  system  was  used  for  a  bridge  at  Prague  (1868)  and  for  the 
Albert  Bridge  in  London  (1872).  It  proved  to  be  uneconomical 
and  unsatisfactory.  A  modified  form,  known  as  the  Gisclard 
System,  was  devised  for  a  bridge  at  Villefranche  in  1908  and 
has  since  been  copied  for  several  other  spans  in  France,  despite 
its  structural  and  aesthetic  drawbacks. 

Practically  all  modern  suspension  bridges  are  stiffened  by 
means  of  a  truss  construction,  either  separate  (Figs.  25-36)  or 
incorporated  in  the  cable  system  (Figs.  37-41).  The  different 
types  of  stiffening  trusses  and  braced-chain  designs  will  be 
discussed  in  separate  sections. 

5.  Methods  of  Lateral  Stiffening.— To  give  the  structure 
lateral  stiffness  against  wind  forces,  the  most  effective  means  is  a 


78 


TYPES  OF  SUSPENSION  BRIDGES 


complete  system  of  lateral  bracing.  If  this  bracing  is  in  the  plane 
of  the  top  or  bottom  chords  of  the  stiffening  truss,  these  chords 
may  act  as  members  of  the  lateral  systems  (Figs.  30,  36); 
otherwise,  separate  wind-chords  must  be  provided  (Figs.  38, 

39>4°)- 

The  wind  bracing  just  described  is  sometimes  supplemented 
by  land-ties  or  wind-anchors,  i.e.,  ropes  connecting  points  on  the 
roadway  to  the  piers  (Fig.  38)  or  to  points  on  shore.  A  hori- 
zontal suspension  system  may  thus  be  formed  (Fig.  38). 

Another  device  for  securing  lateral  stiffness  is  by  building 
the  cables  and  suspenders  in  inclined  planes  (Figs.  27,  30).  This 
"  cradling  "  of  the  cables,  however,  does  not  appreciably  increase 
the  lateral  stability  of  the  structure  if  there  is  but  one  cable  on 
each  side.  If  two  or  more  cables  of  different  inclinations  are 
provided  on  each  side  (Figs.  26,  32),  lateral  stability  is  secured, 
but  at  the  sacrifice  of  equal  division  of  cable  stresses. 

Cradled  cables,  even  if  they  do  not  prevent  lateral  deflection, 
will  help  to  bring  the  resulting  oscillations  more  quickly  to 
rest — an  important  desideratum  in  long  spans. 

6.  Comparison  of  Different  Types  of  Stiffening  Truss. — 
As  a  result  of  a  comparative  estimate  of  different  types  of  stif- 
fened suspension  bridge,  the  following  relative  weights  of  cable 
and  truss  (in  main  span)  were  obtained. 


Relative 

Relative 

Relative 

Type 

Weight 

Weight 

Combined 

of  Cable 

of  Truss 

Weight 

05 

103 

103 

103 

IS 

in 

107 

109 

2S 

IOO 

IOO 

IOO 

2F 

IOI 

89 

95 

3F 

102 

82 

92 

The  hingeless  type  (05)  (Fig.  1 7) ,  gives  the  most  rigid  struc- 
ture, as  a  result  of  the  continuity  of  the  stiffening  truss  over  the 
towers.  The  deflections  will  be  about  f  less  than  those  of  a 
two-hinged  stiffening  truss  of  the  same  dimensions.  This 


COMPARISON  OF  DIFFERENT  TYPES  79 

greater  rigidity  is  secured  at  an  expense  of  only  3  per  cent  increase 
in  the  cost  of  the  structure. 

The  one-hinged  type  (15)  is  the  least  desirable  construction. 
It  has  the  highest  cable  stresses  and  chord  stresses  of  any  type 
of  stiffening  truss.  It  will  cost  more  than  any  other  type;  and 
the  large  variation  in  chord  stresses,  the  abrupt  reversals  of 
shear,  and  the  lack  of  rigidity  are  serious  disadvantages. 

The  two-hinged  types  (2S  and  2F)  are  widely  used  and  are 
probably  the  most  efficient  types,  all  things  considered.  They 
are  more  economical  than  the  continuous  types,  and  are  simpler 
to  figure.  They  are  far  more  rigid  than  type  3F.  The  hinges 
are  located  in  the  towers,  where  they  are  least  objectionable. 

Comparing  types  25  and  2F,  we  find  that  leaving  the  side 
spans  free  (straight  backstays,  Type  2F)  (Fig.  31)  reduces  the 
bending  moments  in  the  main  span.  The  main-span  truss 
weight  is  thus  reduced  by  about  n  per  cent,  without  sensibly 
affecting  the  cable  weight.  For  lightness  of  truss,  type  2F  is 
exceeded  only  by  the  three-hinged  suspension  bridge.  Type  2F 
is  also  more  rigid  than  type  25. 

Suspending  the  side  spans  (Type  25)  (Figs.  30,  35)  makes  the 
cable  more  flexible,  thus  throwing  more  load  on  the  stiffening 
truss.  As  a  result,  about  n  per  cent  is  added  to  the  weight  of 
the  truss  in  the  main  span,  and  the  cable  stress  is  slightly  relieved. 
The  increase  in  cost  of  the  main  span  is  generally  more  than 
offset,  however,  by  the  saving  in  the  side  spans  as  a  result  of 
their  suspension.  Without  any  addition  to  its  weight,  the  cable 
relieves  the  side  spans  of  their  full  dead  load  and  nearly  all  of 
their  live  load.  Type  25  will  consequently  be  more  economical 
than  2F  or  3F  unless  the  conditions  at  the  site  are  favorable  to 
cheap,  independent  approach  spans  (Fig.  31).  Another  advan- 
tage of  suspended  side  spans  is  the  dispensing  with  falsework  for 
their  erection  (Fig.  50). 

The  three-hinged  type  (3F)  is  determinate  for  calculations. 
The  addition  of  the  center  hinge  slightly  increases  the  cable 
stress,  but  effects  a  small  reduction  in  weight  of  stiffening  truss. 
This  type  is  little  used  on  account  of  its  lack  of  rigidity  and  other 
disadvantages  arising  from  the  hinge  at  mid-span.  Inter- 


80  TYPES  OF  SUSPENSION  BRIDGES 

mediate  hinges  are  troublesome  and  expensive  details,  parti- 
cularly in  long  spans;  besides  augmenting  the  deflections,  they 
cause  sudden  reversals  of  shear  under  moving  load,  and  consti- 
tute a  point  of  weakness  and  wear  in  the  structure.  There  is 
also  a  large  waste  of  material  in  the  minimum  chord  sections 
near  the  hinges  which,  in  many  cases,  will  offset  the  theoretical 
reduction  in  the  weight  of  the  stiffening  truss.  Furthermore,  a 
center  hinge  conduces  to  a  serious  distortion  of  the  cable  from  the 
ideal  parabolic  form,  with  a  resulting  overloading  of  some  of  the 
hangers.  In  the  case  of  the  Brooklyn  Bridge  (Fig.  25) ,  the  center 
hinge  or  slip  joint  has  caused  excessive  bending  stresses  in  the 
cable  at  that  point,  and  the  breaking  of  the  adjacent  suspenders; 
1 20  suspenders  near  the  hinge  had  to  be  replaced  by  larger  ropes. 

7.  Types  of  Braced-Chain  Bridges. — A  stiffening  construc- 
tion incorporated  in  the  suspension  system  may  be  used  instead 
of  the  straight  stiffening  truss  at  roadway  level.  The  former 
construction,  as  a  rule,  involves  the  use  of  eyebar  chains  instead 
of  wire  cables  (Figs.  37,  38,  40,  41). 

A  braced-chain  suspension  bridge  is  virtually  an  inverted 
arch  in  which  the  ends  are  capable  of  restricted  horizontal  move- 
ments. The  stresses  are  the  same  as  those  in  an  arch,  but  with 
opposite  signs;  the  principal  stress  is  tension,  instead  of  com- 
pression. 

Braced-chain  bridges  may  be  classified  as  to  the  number  of 
hinges  (OB,  Fig.  41;  2B,  Fig.  39;  3B,  Fig.  38);  or  as  to  outline 
of  the  suspension  system  (Parabolic  Top  Chord,  Figs.  37,  39; 
Parabolic  Bottom  Chord,  Fig.  38;  Parabolic  Center  Line,  Fig.  40; 
Parallel  Chords,  Fig.  41). 

If  the  suspension  system  has  a  parabolic  top  chord  and  a 
straight  bottom  chord  (Type  2BH',  Type  3BUH,  Fig.  37)  it 
corresponds  to  a  spandrel  braced  arch.  The  Lambeth  Bridge, 
London,  is  an  example.  The  top  chord,  like  a  cable,  carries  the 
entire  dead  load.  If  the  live  load  is  not  too  great  in  proportion, 
the  top  chord  will  never  have  its  tensile  stresses  reversed;  it 
may  then  be  built  as  a  flexible  cable  (Lambeth  Bridge)  or  chain 
(Frankfort  Bridge,  Fig.  3-7) .  The  bottom  chord  members  suffer 
reversals  of  stress,  hence  they  must  be  built  as  compression 


TYPES  OF  BRACED-CHAIN  BRIDGES  81 

members.  For  erection,  the  diagonals  should  be  omitted  until 
all  the  dead  load  and  one-half  the  live  load  are  on  the  structure 
at  mean  temperature;  this  procedure  will  minimize  the  extreme 
stresses  in  bottom  chord  and  web  members.  The  advantage  of 
making  the  bottom  chord  straight  is  to  save  hangers  and  extra 
wind  chords. 

To  avoid  having  very  long  diagonals  near  the  ends  of  the 
span,  the  bottom  chord  may  be  bent  up  toward  the  towers 
(Type  2B V,  Fig.  39).  This  construction  has  the  advantage  of 
maximum  truss  depth  near  the  quarter-points  where  the  bending 
moments  are  also  a  maximum.  The  main  part  of  the  lower 
chord  remains  at  the  roadway  level,  thereby  saving  hangers  and 
extra  wind  chords  over  that  length. 

If  the  bottom  chord  is  made  parabolic,  it  becomes  the  prin- 
eipal  carrying  member.  This  outline  is  best  adapted  for  three- 
hinged  systems  (Type  3BL).  A  notable  example  is  the  Point 
Bridge  at  Pittsburgh  (Fig.  38).  In  this  structure,  the  top  chord 
consists  of  two  straight  segments,  intersecting  the  bottom  chord 
at  ends  and  center.  Since  the  bottom  chord  is  the  equilibrium 
curve  for  dead  load,  there  are  no  dead-load  stresses  in  the  top 
chords  or  in  the  web  members.  The  top  chords  must  be  made 
stiff  members,  as  they  are  subject  to  reversals  of  stress.  This 
form  of  suspension  bridge  (Type  3BL)  is  statically  determinate 
and  easily  figured.  It  avoids  the  use  of  long  diagonals  required 
in  the  spandrel  braced  types  (2BV,  Fig.  39;  2BH-,  3BUH,  Fig. 
37),  but  it  requires  the  addition  of  longitudinal  and  lateral  stiffen- 
ing in  the  roadway. 

Instead  of  being  straight  lines  (Fig.  38),  the  two  top-chord 
segments  may  be  curved.  In  a  system  proposed  by  Eads,  they 
are  made  convex  upward. 

To  avoid  reversals  of  stress  in  the  chord  members,  a  form 
known  as  the  Fidler  Truss  may  be  used.  In  this  form  (Type 
35C),  both  chords  are  concave  upward;  and  the  line  midway 
between  top  and  bottom  chords  is  made  parabolic,  so  that  the 
two  chords  will  have  equal  tensions  under  dead  load  and  uni- 
form live  load.  An  example  of  this  form  is  LindenthaPs  Second 
Quebec  Design  (Fig.  40) .  The  outlines  of  the  chords  are  obtained 


82  TYPES  OF  SUSPENSION  BRIDGES 

by  superimposing  the  two  equilibrium  curves  for  total  dead  load 
plus  live  load  covering  each  half  of  the  span  in  turn. 

For  two-hinged  systems  (Type  2BF)  a  crescent-shaped  truss 
may  be  used.  The  top  and  bottom  chains  meet  at  common 
supports  on  the  towers,  where  they  are  connected  to  single 
backstays.  There  are  no  examples  of  this  type. 

If  the  top  and  bottom  chains  are  kept  parallel,  we  have 
either  Type  2BP  or  Type  QBP  (Fig.  41),  according  as  the  truss 
bracing  is  interrupted  or  continuous  at  the  tower.  Both  of  these 
types  are  indeterminate,  and  may  involve  some  uncertainty  of 
stress  distribution.  Unless  the  tower  and  anchorage  details  are 
properly  worked  out.  there  is  danger  of  one  of  the  parallel  chains 
becoming  overstressed  or  inoperative.  Examples  of  these  types 
are  Lindenthal's  Seventh  St.  Bridge  at  Pittsburgh  (Type  2BP) 
and  his  Hudson  River  Bridge  design  (Type  QBP,  Fig.  41). 

An  important  advantage  of  the  braced-chain  system  of  con- 
struction over  the  straight  stiffening  truss  is  the  greater  flexi- 
bility of  outline,  with  the  possibility  of  varying  the  truss  depths 
for  maximum  efficiency.  By  having  the  greatest  depth  of  the 
bracing  at  the  quarter  points  of  the  span,  where  the  maximum 
moments  occur,  the  stiffness  of  the  bridge  with  a  given  expendi- 
ture of  material  is  greatly  increased;  and  by  using  a  shallow 
depth  along  the  middle  third  of  the  span,  the  temperature 
stresses  are  reduced. 

The  braced-chain  construction  (Types  257,  2BH,  or  SB  U) 
saves  one  chord  of  the  truss,  as  the  cable  itself  forms  the  upper 
chord. 

Advantages  of  the  suspended  stiffening  truss  (Figs.  25-36) 
are  more  graceful  appearance,  dispensing  with  extra  wind  chords, 
lower  elevation  of  surfaces  exposed  to  wind,  less  live-load  effect  on 
hangers  and  cables,  simpler  connections,  easier  and  safer  erection. 

In  addition,  the  braced-chain  and  suspended-truss  types 
carry  with  them  the  respective  advantages  of  eyebars  and  wire 
cables,  unless  the  practically  untried  combination  of  overhead 
bracing  with  wire  cables  is  adopted. 

8.  Economic  Proportions  for  Suspension  Bridges. — The 
minimum  ratio  of  side  spans  to  main  span  is  about  J  for  straight 


ECONOMIC   PROPORTIONS    FOR   SUSPENSION   BRIDGES       83 

backstays,  and  about  \  for  suspended  side  spans.  Shorter 
ratios  tend  to  make  the  stresses  or  sections  in  the  backstays 
greater  than  in  the  main  cable.  The  length  of  side  span  is  also 
controlled  by  existing  shore  conditions,  such  as  relative  eleva- 
tions and  suitable  anchorage  sites. 

The  economic  ratio  of  sag  to  span  of  the  cable  between 
towers  is  about  i  if  the  backstays  are  straight  and  about  f  if  the 
side  spans  are  suspended.  (See  the  author's  book  "  Suspension 
Bridges  and  Cantilevers.")  For  light  highway  and  foot-bridges, 
the  sag-ratio  may  be  made  as  low  as  iV  to  iV . 

For  adequate  lateral  stiffness  the  width,  center  to  center, 
of  outer  stiffening  trusses  should  not  be  less  than  about  u^th  of 
the  span. 

The  economic  depth  of  stiffening  truss  is  about  Ath  of  the 
span  (Fig.  31);  although  a  shallower  depth  (Fig.  35),  desirable 
for  aesthetic  reasons,  will  not  materially  augment  the  cost. 
For  a  railroad  bridge,  the  truss  depth  (at  the  quarter  points) 
should  not  be  less  than  about  ^Vth  of  the  span,  or  the  deflection 
gradients  will  exceed  i  per  cent.  (See  "  Suspension  Bridges 
and  Cantilevers.")  For  highway  bridges,  the  depth  may  be 
made  as  low  as  ^oth  to  Ath  of  the  span. 

The  economic  span-limit  for  suspension  bridges  is  about 
3200  feet  (Fig.  41).  For  greater  span-lengths,  the  necessary 
outlay  would  not  be  warranted  by  traffic  returns;  but  there  are 
other  returns,  such  as  civic  development  and  increase  in  realty 
values,  to  justify  longer  spans.  Spans  up  to  $000  feet  may  be 
regarded  as  feasible. 

9.  Arrangements  of  Cross-sections. — The  unit  of  suspension 
bridge  design  is  the  vertical  suspension  system,  consisting  of  a 
cable  (or  group  of  cables)  and  the  corresponding  suspenders  in  a 
vertical  (or  slightly  inclined)  plane. 

Since  the  suspension  systems  are  above  the  roadway,  their 
number  is  limited;  they  seldom  exceed  two  (Figs.  27,  30,  32, 
37-41).  In  wide  bridges  having  a  number  of  roadways,  four 
suspension  systems  may  be  provided  (Figs.  26,  36). 

The  main  carrying  element  in  each  suspension  system  may  be 
a  single  cable  (Figs.  26, 36),  two  cables  side  by  side  (Figs.  27,32), 


84  DETAILS  OF  CONSTRUCTION 

two  cables  superimposed,  or  a  group  of  cables  or  wire  ropes 
(Figs.  27,  29,  30) ;  or  it  may  consist  of  a  single  chain  of  bars  (Figs. 
28,  37),  two  chains  simply  superimposed  (Figs.  33,  34,  39),  or 
two  chains  connected  by  web  members  to  make  a  vertical  stiff- 
ening system  (Figs.  38,  40,  41). 

There  is  generally  one  stiffening  truss  for  each  suspension 
system,  and  in  the  same  plane;  hence,  there  are  ordinarily  two 
(Figs.  30, 32, 37-41),  and  at  most  four  stiffening  trusses  (Fig.  36). 
An  exception  is  the  Brooklyn  Bridge  (Fig.  26),  having  six  stiff- 
ening trusses  for  four  cables;  this,  however,  has  proved  to  be  an 
unsatisfactory  and  inefficient  arrangement. 

Between  the  stiffening  trusses  are  the  roadways,  generally 
on  a  single  deck  (Figs.  27,  30,  33,  34,  37-40).  Sometimes  two 
decks  are  provided,  in  order  to  provide  the  required  number  of 
traffic- ways  (Figs.  26, 32, 36, 41).  Where  two  decks  are  used,  the 
railways  are  best  placed  below  and  the  vehicular  roadways 
above  (Figs.  15,  41).  In  the  Williamsburg  Bridge  (Fig.  32),  a 
transverse  truss  is  employed  to  carry  the  inside  floorbeam 
reactions  to  the  two  outside  suspension  systems. 

The  floorbeams  either  terminate  at  their  connections  to  the 
outer  suspension  systems  (Figs.  26,  27,  30,  37-40);  or  they 
extend  beyond  as  cantilever  brackets  to  carry  outside  sidewalks 
or  roadways  (Figs.  32,  36,  41).  The  latter  arrangement  saves 
floorbeam  weight;  reduces  width  of  towers,  piers  and  anchorages; 
and  helps  in  the  separation  of  different  traffic-ways.  In  very 
long  spans,  the  first  arrangement  (with  all  roadways  inside) 
may  be  necessary  in  order  to  maintain  the  requisite  width  between 
trusses  for  lateral  stiffness. 

Where  there  are  four  suspension  systems,  the  floorbeams 
may  be  made  continuous  for  greater  stiffness  (Fig.  36) ;  or  they 
may  be  provided  with  hinges  to  eliminate  the  indeterminateness. 

10.  Materials  used  in  Suspension  Bridges. — The  stiffening 
trusses  are  generally  built  of  structural  steel,  but  nickel  steel  or 
other  alloy  steels  may  be  used;  and  for  minor  structures,  timber 
trusses  have  been  employed. 

The  cables  are  generally  made  of  galvanized  steel  wires 
having  an  ultimate  strength  of  200,000  to  230,000  pounds  per 


WIRE  ROPES  85 

square  inch,  and  an  elastic  limit  as  high  as  150,000  pounds  per 
square  inch. 

The  suspenders  are  generally  galvanized  steel  ropes.  These  are 
manufactured  in  diameters  ranging  from  ij  to  2\  inches,  and 
have  a  tested  ultimate  strength  given  by  80,000  X  (diameter)2. 

In  smaller  bridges  (Figs.  29,  30),  the  cables  may  be  made  of 
these  galvanized  steel  ropes  instead  of  parallel  wires. 

The  towers  are  generally  built  of  structural  steel  (Figs.  30,  31, 
35>  37>  38>  4i);  although  stone  (Figs.  25,  27,  29,  33),  concrete, 
and  timber  have  been  used. 

Cast  steel  is  used  for  all  castings,  such  as  saddles  (Figs.  32,  33), 
cable  bands  (Figs.  32,  36),  strand  shoes  (Figs.  32,  36),  anchorage 
knuckles  (Figs.  32,  33),  and  anchor  shoes  (Figs.  33,  37,  38).  Sus- 
pender sockets  (Figs.  32,  36)  are  made  by  drop-forging  and 
machining. 

If  chains  are  adopted  instead  of  wire  cables,  alloy  steels  may 
be  advantageously  employed.  In  a  competition  for  a  suspen- 
sion bridge  at  Worms,  the  Krupp  firm  guaranteed  nickel  steel 
eyebars  with  an  ultimate  strength  of  100,000  to  120,000,  an 
elastic  limit  of  70,000  pounds  per  square  inch,  and  an  elongation 
of  15  per  cent.  For  Lindenthal's  Manhattan  Bridge  design 
(1902),  nickel  steel  eyebars  with  an  ultimate  strength  of  100,000 
and  with  20  per  cent  elongation  were  to  be  used.  The  chains 
for  the  Elizabeth  bridge  at  Budapest  (Fig.  34)  were  made  of  open- 
hearth  steel  with  70,000  to  80,000  ultimate  strength  and  with 
20  per  cent  elongation. 

Under  average  conditions,  the  substitution  of  nickel  steel 
affords  a  saving  of  10  to  15  per  cent  in  the  cost  of  a  chain  or  a 
stiffening  truss. 

11.  Wire  Ropes. — Galvanized  steel  ropes  used  for  suspenders 
and  for  small  bridge  cables  (Figs.  29,  30),  are  manufactured  in 
diameters  ranging  from  ij  to  2f  inches.  Each  rope  consists  of 
7  strands,  each  strand  containing  7,  19,  37  or  6 1  wires;  the  wires 
are  twisted  into  strands,  in  the  opposite  direction  to  the  twist  of 
the  strands  into  rope,  the  angle  of  twist  being  about  18°.  The 
weight  of  the  rope  in  pounds  per  lineal  foot  is  I.68X  (diameter)2. 

The  strength  of  a  twisted  wire  rope  is  less  than  the  aggregate 


DETAILS  OF  CONSTRUCTION 


strength  of  the  indi- 
vidual wires.  The  spi- 
ral wires  are  stressed 
about  4  or  5  per  cent 
higher  than  the  mean 
stress  per  square  inch 
in  the  rope,  and  the 
center  wire  is  stressed 
15  per  cent  higher 
than  the  spiral  wires. 
The  tested  ultimate 
strength  of  galvanized 
steel  suspension  bridge 
rope  is  given  by  80,000 
X  (diameter)2. 

When  twisted  wire 
ropes  are  used  for 
cables,  care  must  be 
observed,  when  apply- 
ing the  fundamental 
design  formulas,  to 
allow  for  the  reduced 
elastic  coefficient  (E) 
of  this  material;  it  is 
only  about  f  of  the 
value  of  E  for  struc- 
tural  steel. 

The  coefficient  of 
elasticity  (E)  of  a  single 
rope  strand  with  an 
angle  of  twist  of  18°  is 
85  per  cent  of  E  for  par- 
allel wires,  or  about 
24,000,000.  The  co- 
efficient of  elasticity 
(E)  of  a  twisted  wire 
rope  composed  of  7  or 


WIRE  ROPES  87 

more  strands  is  85  per  cent  of  E  for  a  single  strand,  or  about 

20,000,000. 

Twisted  wire  ropes  have  a  large  initial  stretch  under  load, 
on  account  of  the  spiral  lay  of  the  wires  and  strands.  Conse- 
quently, at  small  loads,  tests  show  a  high  rate  of  stretch  yielding 
a  modulus  of  elasticity  (E)  as  low  as  10,000,000.  After  the 
initial  stretch  has  been  taken  up  (at  a  unit  stress  of  about  20,000 
pounds  per  square  inch) ,  the  rate  of  elongation  is  considerably  re- 
duced, yielding  a  value  of  20,000,000  for  the  true  elastic  coefficient 
(E).  The  lower  values  of  E  (10,000,000  to  15,000,000)  are  to  be 
used  in  estimating  the  dead-load  elongation  of  the  cable  (if  com- 
posed of  wire  ropes),  and  the  higher  value  (20,000,000)  should 
be  used  in  figuring  live-load  and  temperature  stresses. 

On  account  of  the  high  and  variable  elongations,  including 
the  influence  of  time,  suspenders  and  cables  made  of  wire  ropes 
should  preferably  be  provided  with  screw  and  nut  adjustments  to 
regulate  their  lengths  to  the  assumed  deflections  and  elevations. 

Cables  may  be  built  either  of  twisted  wire  ropes  or  of  parallel 
wires.  For  long  or  heavy  spans,  parallel  wire  construction  is 
best  adapted;  for  light  bridges,  the  use  of  twisted  wire  ropes 
may  be  more  convenient  and  expeditious. 

In  cables  formed  of  twisted  wire  ropes,  the  individual  ropes 
are  limited  to  250  to  300  wires  each,  so  as  to  avoid  excessive 
stiffness  and  difficulty  of  handling;  consequently,  large  cable 
sections  require  several  such  ropes. 

A  multi-strand  cable  •  may  be  formed  of  twisted  strands 
surrounding  a  straight  central  strand;  or  of  parallel  strands 
united  at  intervals  by  clamps.  Twisted  strands  ensure  a  more 
even  division  of  load,  except  that  the  central  strand  carries  a 
little  more  than  its  share;  but  the  resulting  cable  suffers  greater 
elongation  under  load.  Moreover,  since  a  twisted-strand  cable 
must  be  erected  as  a  unit,  it  is  limited  in  weight  and  section. 
Equal  stressing  of  parallel  strands  is  dependent  upon  the  effi- 
ciency of  the  clamps  or  bands  in  gripping  them.  An  advantage 
of  the  parallel  construction  with  bolted  clamps  is  the  ease  of 
correcting  overstress  in  individual  strands  and  of  replacing 
damaged  strands.  Clamping  systems  have  been  designed  for 


88 


DETAILS  OF  CONSTRUCTION 


large  groups  of  parallel  ropes,  to  ensure  unit  stress  action  and  to 
facilitate  renewal  of  individual  ropes;  at  the  same  time  pre- 
serving ample  spacing  between  the  ropes  to  permit  inspection 
and  protection  against  rusting. 


A  twisted  wire  cable  of  patent  locked  wire  has  been  developed. 
In  it  the  spiral  wires  have  trapezoidal  and  Z-sections,  locking 
together  so  as  to  leave  practically  no  voids.  The  advantages 


PARALLEL  WIRE  CABLES  89 

are  compactness,  smooth  outer  surface,  firm  gripping  of  the 
individual  wires,  and  sealing  against  the  entrance  of  moisture. 
Cables  of  this  construction,  ready  to  erect,  have  been  made  in 
single  strands  up  to  800  tons  tensile  strength,  and  in  seven 
strands  up  to  1500  tons. 

The  application  of  twisted  ropes  on  a  large  scale  involves 
problems  requiring  further  study;  whereas  parallel  wire  con- 
struction has  had  ample  satisfactory  demonstration  in  the 
largest  existing  suspension  bridges. 

12.  Parallel  Wire  Cables. — Parallel  wire  cables  have  the 
advantages  of  maximum  compactness,  maximum  uniformity  of 
stress  in  all  the  wires,  and  the  easiest  and  safest  connection  of 
the  cable  to  the  anchorage.  Twisted  wire  ropes  are  used  for 
shorter  spans,  up  to  600  or  700  feet,  to  save  time  in  erection. 
Parallel  wires  are  applicable  to  spans  of  any  length,  and  will 
cost  somewhat  less  than  twisted  ropes  of  the  same  strength; 
they  will  not  stretch  as  much  as  twisted  ropes,  and  will  there- 
fore keep  more  of  the  load  off  the  stiffening  truss.  The  only 
disadvantage  of  parallel  wire  cables  is  that  they  consume  several 
weeks  or  months  in  erection. 

A  common  size  of  wire  for  cables  is  No.  6  (Roebling  gauge) 
which  is  0.192  inch  diameter  and  weighs  0.0973  pound  per  foot 
before  galvanizing;  after  galvanizing,  the  diameter  is  about 
0.195  inch?  and  the  weight  is  practically  TO  pound  per  foot. 
The  breaking  strength  of  this  wire  at  220,000  pounds  per  square 
inch  is  6400  pounds;  the  elastic  limit  at  150,000  pounds  per 
square  inch  is  4350  pounds;  the  working  stress  at  75,000  pounds 
per  square  inch  is  2180  pounds  per  single  wire.  Other  common 
sizes  of  wire  for  cables  are  No.  7  (0.177  inch  diameter)  and  No.  8 
(0.162  inch  diameter),  recommended  for  shorter  spans. 

About  250  to  350  of  these  wires  are  treated  as  a  single  strand 
during  erection.  The  cable  consists  of  7,  19,  37  or  61  of  these 
strands.  At  the  anchorages,  the  strands  are  looped  around 
grooved  shoes  (Fig.  36)  which  are  pin-connected  to  the  anchorage 
eyebars  (Fig.  32).  For  the  rest  of  their  length,  the  strands  are 
compacted  and  bound  to  form  a  cylindrical  cable  of  parallel  wires. 

For  security  agamst  corrosion,  the  wire  should  be  galvanized. 


90  DETAILS  OF  CONSTRUCTION 

The  only  drawback  is  a  reduction  of  about  7  per  cent  in  the 
strength  of  the  wire  per  square  inch  of  final  gross  section  (4  per 
cent  actual  reduction  in  the  strength  of  the  wire,  and  3  per  cent 
increase  in  gross  section.) 

The  splicing  of  individual  wires  was  formerly  effected  by 
wrapping  the  overlapping  ends  with  fine  wire.  A  more  efficient 
splice  (giving  95  per  cent  efficiency)  is  made  by  mitering  the 
ends,  threading  them  and  connecting  with  small  sleeve-nuts 
(Figs.  32,  36).  Both  methods  have  the  disadvantage  of  disturb- 
ing the  uniformity  of  the  cable  section.  To  reduce  the  number 
of  such  splices,  the  lengths  of  the  individual  wirer,  as  manu- 
factured have  been  increased  to  3300  feet.  In  some  French 
bridges,  the  ends  of  the  wires,  after  beveling,  were  joined  by  solder- 
ing; but  the  heat  reduces  the  strength  of  the  wire  at  the  splice. 

Besides  using  galvanized  wires,  additional  protection  is 
secured  by  providing  a  tight  and  continuous  wire  wrapping 
around  the  cable.  Soft,  annealed,  galvanized  wire  of  No.  8  or 
No.  9  Roebling  gauge  is  commonly  used.  The  function  of  this 
wrapping  is  to  exclude  moisture,  to  protect  the  outer  wires,  and 
to  hold  the  entire  mass  of  wires  so  tightly  as  to  prevent  chafing 
and  ensure  united  stress  action. 

No  record  can  be  found  of  any  rusting  of  wire  cables  employ- 
ing either  or  both  of  the  above  described  methods  of  protection. 

13.  Cradling  of  the  Cables. — In  the  majority  of  suspension 
bridges,  the  main  span  cables  do  not  hang  vertically  but  in 
planes  inclined  toward  one  another,  the  inclination  ranging  from 
i  :  20  to  as  much  as  i  :  6.  The  stiffening  trusses,  however,  are 
kept  vertical.  Even  in  designs  with  overhead  bracing,  the 
suspension  systems  have  been  cradled  with  inclinations  ranging 
from  i  :  20  to  i  :  16. 

Cradling  is  employed  principally  because  it  is  supposed  to 
augment  the  lateral  stiffness  of  the  structure ;  however,  the  advan- 
tage in  this  respect  over  vertical  cables  is  but  slight.  With  an  in- 
clination of  i  :  10,  the  increased  resistance  to  lateral  displacement 
is  only  i  per  cent.  Moreover,  with  cradled  cables  any  lateral  dis- 
placement is  accompanied  by  a  tilting  of  the  suspended  structure, 
resulting  in  secondary  stresses  which  are  difficult  to  evaluate. 


CRADLING  OF  THE  CABLES 


91 


The  following  table  gives  data  on  the  wire  cables  of  the 
East  River  suspension  bridges : 


/ 

Brooklyn 
(Fig.  25) 

Williamsburg 
(Fig.  31) 

Manhattan 
(Fig.  35) 

Date 

1876-1883 

l8o8—  IQO3 

IQO^—  IOOQ 

Main  span  

1595.5  ft- 

1600  ft. 

1470  ft. 

Cable-sag      .        

128  ft. 

177  ft. 

1  60  ft. 

Total  load  p  1.  f 

^.soo  Ib. 

7^,000  lb. 

104,000  lb. 

Number  of  cables  
Strands  per  cable  
Wires  per  strand 

4 
19 
278 

4 
37 
208 

4 
37 
256 

Wire  diameter 

165  in. 

192  in. 

inr  in. 

Total  cross-section  
Cable  diameter 

533  sq.  in. 
I^T  in. 

888  sq.  in. 
i8f  in. 

1092  sq.  in. 
21  j  in. 

Size  of  wrapping  wire 

I31?  in. 

148  in 

Max.  stress  in  cables.  .  .  . 
Ult.  strength  of  cables..  . 

47,500  Ib./sq.  in. 
160,000  Ib./sq.  in. 

50,300  Ib./sq.  in. 
200,000  Ib./sq.  in. 

73,000  Ib./sq.  in. 
210,000  Ib./sq.  in. 

Resistance  to  lateral  displacement  is  more  significantly 
improved  in  proportion  as  the  cable-sag  is  reduced  and  as  the 
weight  of  the  suspended  structure  is  increased. 

If  cradling  is  adopted,  the  cables  should  not  be  wrapped  until 
they  are  pulled  into  the  final  inclined  position;  otherwise  serious 
local  stresses  will  be  produced  in  the  cable  wires  near  the  saddles. 

If  the  cables  are  cradled,  the  saddle  reactions  on  the  towers 
will  be  correspondingly  inclined  unless  the  backstay  cables  are 
made  divergent;  the  latter  arrangement  (used  in  the  Williamsburg 
Bridge,  Fig.  32)  increases  the  necessary  width  of  the  anchorage. 

Cradling  will  be  effective  in  producing  lateral  stiffness  if  two 
cables  of  different  inclination  are  provided  on  each  side  (Figs.  26 
and  32).  This  arrangement  has  the  disadvantage  of  throwing 
unequal  load  on  the  cables  when  the  wind  acts  on  the  structure; 
load  is  added  to  the  cables  inclined  in  the  direction  of  the  wind, 
and  those  inclined  in  the  opposite  direction  are  relieved  of  load. 

14.  Anchoring  of  the  Cables. — Parallel  wire  cables  are  an- 
chored by  making  the  end  of  each  strand  in  the  form  of  a  sling. 
With  the  wrapping  omitted  at  the  end  of  the  cable,  the  free  wires 
loop  around  a  half-round,  flanged  casting  called  a  strand  shoe 


92 


DETAILS  OF  CONSTRUCTION 


CONNECTION  OF  CABLE  TO  ANCHOR  CHAIN 


ANCHORAGE  CABLE  BAND  AND  SUSPENDERS 

FIG.  32. — Williamsburg  Bridge. 
(Type  2/0. 


ANCHORING  OF  THE  CABLES  93 

(Fig.  36),  and  then  pass  back  into  the  strand.  Large  cables  are 
divided  for  this  purpose  into  7,  19  or  37  strands;  and  such 
cables  accordingly  have  7,  19,  or  37  strand  shoes  at  each  end. 
Steel  pins  pass  through  these  shoes  for  connection  to  the  anchor- 
age eyebars  (Fig.  32). 

The  strand  shoes  are  grouped  into  a  number  of  horizontal 
rows  (generally  2  or  4),  and  the  anchor  chain  divides  into  an 
equal  number  of  branches  to  effect  the  connection  (Fig.  3  2) . 

About  10  feet  forward  of  the  shoe,  the  two  halves  of  a  strand 
are  combined  into  one;  and  all  strands,  before  leaving  the 
masonry,  are  squeezed  into  a  round  cable. 

The  shoes  have  slotted  pin-holes  which  are  provided  with 
shim-blocks  (Fig.  36)  to  permit  regulation  of  the  individual 
strands  before  combining  into  a  cable. 

Bending  the  wires  around  the  shoe  produces  bending  stresses 
exceeding  the  elastic  limit;  but  the  resulting  stretching  of  the 
outer  fibers  redistributes  the  stress  over  the  cross-section  of  the 
wire;  with  properly  ductile  steel  wire,  the  strength  at  the  loop 
is  not  materially  impaired. 

If  the  wire  is  very  hard,  or  if  the  cable  consists  of  wire  ropes, 
a  larger  radius  of  curvature  must  be  provided  or  some  other 
form  of  connection  must  be  used. 

For  wire  ropes  a  larger  shoe  is  used,  with  the  end  of  the  rope 
fastening  into  a  socket  after  bending  around  the  shoe. 

The  sling  construction  is  avoided  by  setting  the  ends  of  the 
strands  or  ropes  directly  into  steel  sockets.  After  inserting  the 
rope  into  the  expanding  bore  of  the  socket,  the  wires  are  pried 
apart  and  spread  with  a  point  tool,  and  the  intervening  space 
filled  with  fusible  metal  (preferably  molten  zinc).  Such  socket 
connections  are  now  made  to  develop  the  full  strength  of  the 
rope.  The  sockets  may  be  designed  to  bear  directly  against 
the  under  side  of  the  anchor  girder;  or  they  may  be  threaded 
to  receive  the  end  of  a  rod  which  serves  as  a  continuation  of  the 
strand. 

15.  Construction  of  Chains. — Chains  may  be  constructed  of 
horizontal  flats  piled  together  and  spliced  at  intervals  by  means 
of  friction  clamps  with  bolted  flanges.  Suspenders  are  bolted 


CONSTRUCTION  OF  CHAINS 


to  these  clamps.  This  laminated 
construction  is  subject  to  high 
secondary  stresses  from  bending. 

Chains  may  be  constructed  of 
closed  links  overlapping  around 
connecting  pins  to  which  the 
hangers  are  attached  (Fig.  39). 

Chains  may  consist  of  eyebars 
or  flats  bored  at  their  ends  to 
receive  pins  (Figs.  33,  34,  38,  40). 
Generally,  single-pin  connections 
are  used,  and  the  number  of  bars 
alternates  in  successive  panels. 
Otherwise,  short  two-pin  connect- 
ing bars  may  be  used,  permitting 
the  number  of  eyebars  to  be  the 
same  in  adjacent  panels. 

In  American  practice  (Figs.  38, 
40),  forged  eyebars  are  used.  In 
European  practice  (Figs.  17,  33, 
34) ,  the  eyebars  are  made  by  weld- 
ing or  riveting,  or  by  cutting  from 
wide  flats.  The  last  is  an  extrava- 
gant procedure. 

Where  flats  are  used,  the  re- 
duction of  section  by  the  pin- 
holes  may  be  largely  made  up  by 
riveting  pin-plates  at  the  ends  of 
the  bars. 

Chains  composed  of  vertical 
flats  riveted  together  have  been 
proposed,  but  the  secondary 
stresses  from  bending  would  be 
very  high. 

For  long  spans,  the  chains 
would  have  large  cross-sections, 
requiring  pins  of  excessive  length. 


w 


I 


96  DETAILS  OF  CONSTRUCTION 

This  is  circumvented  by  using  two  chains,  either  side  by  side 
or  superimposed;  in  the  latter  case,  if  the  panels  are  not  too 
short,  successive  hangers  are  connected  alternately  to  the  upper 
and  lower  chains  (Figs.  33,  34). 

A  disadvantage  of  chain  construction  is  unequal  division  of 
stress  in  the  individual  bars  between  two  pins.  This  may  be 
caused  by  inaccuracies  in  length,  differences  in  temperature, 
variations  in  elastic  modulus,  bending  of  the  pins,  and  eccentric 
suspender  loading.  The  unequal  stressing  of  the  eyebars  is 
frequently  apparent  on  superficial  examination  or  upon  com- 
paring the  ringing  pitch  under  hammer  blows.  Actual  measure- 
ments (by  comparing  deflections  under  lateral  test  loads)  have 
revealed  varying  stresses  in  a  single  group  of  eyebars  ranging 
from  40  to  200  per  cent  of  the  mean  stress. 

16.  Suspender  Connections. — Cable  Bands  and  Sockets.— 
The  attachment  of  the  suspenders  to  the  cable  is  generally  made 
by  means  of  cast  steel  collars  called  cable  bands  (Figs.  26,  32,  36). 
The  cable  band  may  be  an  open  ring  with  flanged  ends  to  receive 
a  clamping  and  connecting  bolt  (Fig.  26) .  More  generally  it  is 
made  in  two  halves  with  flanges  (Figs.  32,  36).  The  band  must 
grip  securely  to  prevent  slipping.  The  inside  of  the  band  should 
be  left  rough  to  minimize  the  tendency  to  slip  on  the  cable; 
and  space  should  be  left  between  the  flanges  for  taking  up  any 
looseness  of  grip,  when  necessary.  A  cam-clamping  device  has 
been  proposed  for  automatically  increasing  the  grip  as  load  is 
applied  through  the  suspender. 

If  the  hangers  are  of  rigid  section,  they  are  bolted  to  vertical 
flanges  cast  integral  with  the  cable  band  for  this  purpose. 

If  rope  suspenders  are  used,  the  cable  band  is  cast  with  a 
groove  or  saddle  to  receive  the  rope  which  passes  over  it  (Figs.  32, 
36).  Oh  account  of  the  varying  slope  of  the  cable,  the  grooves 
in  the  cable  bands  are  at  varying  angles,  requiring  a  number  of 
different  patterns.  To  avoid  this,  the  bearing  flange  of  the 
grooves  may  be  made  curved  in  elevation. 

If  the  cable  is  used  as  a  chord  of  an  overhead  bracing  system, 
the  rigid  web  members  connect  to  the  cable  bands;  and  the 
latter  must  be  made  long  enough,  with  ample  clamping  bolts,  to 


SUSPENDER  CONNECTIONS 


97 


98  DETAILS  OF  CONSTRUCTION 

develop  the  friction  requisite  to  take  up  the  chord  increment  of 
the  web  stresses.  A  tight  layer  of  wire  wrapping  against  the 
ends  of  the  cable  bands  will  add  to  their  security  against  slipping. 

The  frictional  grip  (in  pounds)  attainable  in  a  cable  band, 
with  maximum  permissible  stress  in  the  bolts,  is  70  to  ioond2, 
where  n  is  the  number  and  d  is  the  diameter  (in  inches)  of  the 
clamping  bolts.  By  this  relation  may  be  determined  the  number 
of  bolts  required  to  resist  a  given  component  parallel  to  the  cable. 

If  the  cable  consists  of  a  cluster  of  wire  ropes,  soft  metal 
fillers  should  be  inserted  within  the  band  to  improve  the  grip 
and  to  exclude  moisture. 

The  cable  band  should  be  designed  so  as  to  prevent  the  admis- 
sion of  moisture  to  the  cable.  The  flanges  should  be  designed 
for  excluding  rain,  and  the  joints  all  around  should  be  securely 
calked.  The  band  should  preferably  be  undercut  at  both  ends 
for  the  insertion  of  the  first  few  turns  of  the  wire  wrapping 

(Fig.  36). 

The  free  ends  of  the  suspender  ropes  are  secured  in  sockets 
made  of  high-grade  steel  drop-forgings  (Figs.  32,  36).  The  end 
of  the  rope  is  inserted  into  the  expanding  shell  of  the  socket,  the 
ends  of  the  wires  are  spread  apart  and  the  interstices  are  filled 
with  molten  metal  (preferably  zinc)  which  will  not  shrink 
appreciably  on  cooling.  This  fastening  of  the  end  of  the  rope 
is  found,  by  test,  to  be  unaffected  by  the  ultimate  loads  causing 
failure  of  the  rope. 

Closed  sockets  terminate  in  a  closed  loop  with  which  other 
links  can  be  engaged.  Open  sockets  terminate  in  two  parallel 
eye-ends  to  receive  a  bolt  or  pin  for  connection  to  other  struc- 
tural parts.  Threaded  sockets  (Figs.  32,  36)  are  cylindrical  and 
are  threaded  on  the  outside  to  receive  adjusting  and  holding 
nuts ;  these  sockets  may  be  passed  through  truss  chords  or  girder 
flanges,  with  the  nuts  bearing  up  against  the  lower  cover  plates 
of  these  members  (Fig.  36). 

Sockets  are  furnished  by  the  wire  rope  manufacturers,  either 
loose  or  fastened  to  the  ropes. 

17.  Suspension  of  the  Roadway. — The  suspenders  may  con- 
sist of  wire  ropes  (Figs.  26-28,  30-32,  36);  or  of  rods,  bars  or 


SUSPENSION  OF  THE  ROADWAY 


99 


ELEVATION  AND  PLAN 


illiT  block      SEC.  A-A 

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Cast  Steel  Strand  Shoe 


Cable  Band 
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CABLE  BAND 


SUSPENDERS 
AND  CONNECTION 


FIG.  36. — Manhattan  Bridge. 
(Type  25). 


100  DETAILS  OF  CONSTRUCTION 

rolled  shapes  (Figs.  29,  33,  34,  38-41).  There  may  be  one  (Fig. 
26),  two  (Fig.  30),  or  four  suspenders  (Figs.  32,36)  at  a  panel  point. 
If  the  hangers  are  made  of  rigid  rods  (instead  of  wire  ropes), 
bending  stresses  due  to  lateral  or  longitudinal  swaying  of  the 
bridge  are  avoided  by  inserting  pin  connections  or  links  (Figs.  38, 

39>  4o). 

Solid  steel  rods  used  for  hangers  generally  have  a  high  slender- 
ness  ratio  and  are  subject  to  bending  and  to  vibrations;  to 
provide  greater  stiffness,  tubular  and  built-up  sections  have  been 
substituted  (Figs.  34,  38,  41). 

Where  there  are  two  or  more  suspenders  at  a  panel  point, 
the  possibility  of  unequal  division  of  load  should  be  taken  into 
consideration.  Equalizers  may  be  used  to  advantage  (Fig.  32). 

After  passing  around  the  cable  band,  the  suspender  may 
extend  down  as  two  separate  ropes  (Fig.  36) ;  or  the  short  end 
may  be  clamped  to  the  main  suspender,  which  then  extends 
down  as  a  single  rope  (Fig.  32). 

The  suspenders  may  connect  directly  to  the  floorbeams 
(Figs.  26,  27,  29,  38,  41),  or  to  the  top  or  bottom  chord  of  the 
stiffening  truss  (Figs.  32,  36).  IL  the  latter  case,  the  floorbeams 
frame  into  the  chords  or  into  the  posts  of  the  stiffening  truss. 

For  connection  to  the  floorbeams  or  chords,  the  suspenders 
may  pass  through  and  bear  up  against  the  lower  cover  plate 
with  the  aid  of  washers  or  special  castings  (Figs.  26,  32,  36);  or 
they  may  loop  around  the  floorbeams  or  chords  either  directly 
(Fig.  27)  or  with  the  aid  of  steel  cross-pieces  or  yokes. 

Connecting  the  suspenders  to  the  top  chord  of  the  stiffening 
truss  requires  the  entire  length  of  the  cable  to  be  above  the  truss; 
this  has  aesthetic  advantages  (see  Figs.  28,  34,  35),  but  it  adds 
the  depth  of  the  truss  to  the  required  height  of  the  towers. 
Lowering  the  cable  saves  height  (Figs.  25-27,  29-33),  Dut 
requires  either  lengthening  of  the  floorbeams  or  spreading  of  the 
trusses  or  towers. 

Another  method  of  suspending  the  roadway  is  to  loop  the 
suspender  rope  under  a  small  saddle  casting  from  which  there 
extend  downward  rigid  rods  terminating  in  holding  nuts  (Fig.  32) 
or  steel  flats  bored  to  receive  connecting  pins. 


CONSTRUCTION  OF  STIFFENING  TRUSSES  101 

Provision  for  adjustment  of  the  hangers  may  be  made  by 
means  of  the  holding  nuts  (Figs.  32,  36),  or  by  means  of  sleeve 
nuts  or  turn  buckles  with  right  and  left  threads  (such  as  shown 
in  Fig.  32).  Some  engineers  prefer  to  omit  provision  for  adjust- 
ment, depending  upon  careful  computation  of  required  length 
before  cutting  the  ropes  and  attaching  the  sockets. 

18.  Construction  of  Stiffening  Trusses. — The  function  of 
the  stiffening  truss  is  to  limit  the  deformations  of  the  cable  and 
to  so  distribute  any  concentrated,  unsymmetrical,  or  non- 
uniform  loads  as  to  keep  the  suspender  tensions  in  a  constant 
proportion  (or  equal  if  the  cable  is  parabolic).  In  other  words, 
the  stiffening  truss  is  required  to  hold  the  cable  (or  chain)  in  its 
initial  curve  of  equilibrium.  This  will  limit  the  deflections  of  the 
structure,  and  will  resist  the  setting  up  of  vertical  oscillations. 

The  first  suspension  bridge  provided  with  a  stiffening  truss 
was  the  820-foot  railway  span  at  Niagara,  built  by  John  A.  Roeb- 
ling  in  1851-55.  The  Brooklyn  Bridge  (Fig.  26),  completed  in 
1883  by  Roebling's  son,  was  built  with  4  cables  and  6  stiffening 
trusses,  and,  in  addition,  was  provided  with  diagonal  stays. 

Until  comparatively  recent  years,  stiffening  trusses  were 
only  roughly  figured  and  were  made  of  constant  section.  The 
scientific  design  of  suspension  bridges  dates  from  about  1898. 

Stiffening  trusses  are  generally  built  with  parallel  chords 
(Figs.  25,  30,  32,  35);  a  small  variation  in  depth  is  sometimes 
introduced  (Fig.  34).  To  prevent  an  unsightly  and  otherwise 
undesirable  sag  under  load,  and  to  counteract  the  illusion  of 
sag,  a  generous  camber  is  usually  provided  (Fig.  31). 

The  web  system  may  be  of  the  single  Warren  (Figs.  30,  35) 
double  intersection  (Fig.  34),  or  latticed  types  (Fig.  31).  The 
J^-truss  may  also  be  applied  to  advantage.  Instead  of  a  truss, 
plate  girders  may  be  used;  the  Vierendeel  girder  or  quadrangular 
truss  (like  the  floorbeam  in  Fig.  41)  has  also  been  proposed. 

To  make  the  design  statically  determinate,  a  hinge  at  the 
center  of  the  stiffening  truss  is  necessary  (Type  3F  or  35,  Fig.  26) ; 
but  this  construction  has  many  drawbacks.  In  long  spans, 
the  angle  change  at  the  hinge  would  be  so  great  as  to  cause 
serious  bending  stresses  in  the  cable  and  overloading  of  adjacent 


102  DETAILS  OF  CONSTRUCTION 

suspenders.  Moreover,  the  wind  stresses  must  be  transferred 
through  the  hinge,  and  the  details  become  more  difficult  and 
costly. 

Making  the  truss  continuous  past  the  towers  (Types  OF, 
05,  Fig.  34)  yields  more  effective  stiffening:  either  less  material 
is  required  or  the  deflections  are  reduced;  furthermore,  the 
impact  effects  of  moving  loads  entering  the  main  span  are 
reduced.  On  the  other  hand,  continuity  renders  the  structure 
indeterminate  (in  the  third  degree);  inaccuracies  in  construc- 
tion, settlement  of  supports,  and  unequal  warming  of  the  chords 
will  affect  the  stresses  adversely. 

Introducing  hinges  in  the  continuous  stiffening  truss  relieves 
the  indeterminateness  and  the  accompanying  uncertainty  of 
stress  conditions.  In  the  Williamsburg  Bridge  (Fig.  31)  a  hinge 
is  placed  in  each  side  span,  close  to  the  tower;  in  a  prize  design 
for  the  Elizabeth  Bridge  at  Budapest  (Fig.  34),  two  hinges  were 
located  in  the  main  span;  in  both  cases,  the  resulting  system  is 
singly  indeterminate.  In  the  usual  two-hinged  construction 
(Types  2F,  26",  Figs.  15,  36),  the  truss  is  hinged  or  interrupted 
at  the  towers. 

As  in  other  indeterminate  structures,  all  precautions  must  be 
observed  in  construction  to  avoid  false  erection  stresses.  If 
the  suspenders  are  adjustable,  a  definite  apportionment  of  dead 
load  between  cable  and  stiffening  truss  may  be  secured.  The 
stiffening  truss  may  be  totally  relieved  of  dead-load  stress  by 
adjusting  it  under  full  dead  load  and  mean  temperature  to  the 
exact  form  it  had  when  assembled  in  the  shop  at  the  same  tem- 
perature; or  by  omitting  certain  members  until  full  dead  load  (at 
mean  temperature)  is  on  the  structure.  In  any  case,  the  joints 
should  not  be  riveted  until  the  dead  load  is  on  and  all  adjust- 
ments are  made. 

The  stiffening  truss  may  be  made  of  any  height,  depending 
upon  the  degree  of  stiffness  desired.  With  increasing  depth, 
the  stiffness  is  naturally  augmented;  and,  up  to  a  certain  limit, 
material  is  saved  in  the  chords  of  the  stiffening  truss.  Beyond 
this  limit  (economic  depth  =  about  ^Vth  of  the  span),  the  chord 
sections  commence  to  increase  as  a  result  of  the  high  temperature 


BRACED-CHAIN  CONSTRUCTION  103 

stresses.  For  the  sake  of  appearance,  a  somewhat  shallower 
depth  may  be  used  (Fig.  35)  without  materially  affecting  the 
economy.  If  the  truss  has  a  center  hinge,  a  greater  depth 
becomes  economical,  since  temperature  stresses  and  uniform 
load  stresses  practically  vanish.  If  the  deformations  are  to  be 
limited  so  as  not  to  produce  a  deflection  gradient  exceeding 
i  per  cent,  the  depth  must  be  made  not  less  than  Ath  of  the 
span,  whether  two-hinged  or  three-hinged. 

Bearings  must  be  provided  at  the  towers  and  abutments  to 
take  the  positive  reactions  of  the  stiffening  truss  (even  if  continu- 
ous) ;  and  these  points  of  the  truss  must  also  be  anchored  down  to 
resist  the  uplift  or  negative  reactions.  At  the  expansion  bearings, 
the  anchorage  must  be  so  designed  as  to  permit  free  horizontal 
movement;  this  may  be  accomplished  either  by  the  use  of 
anchored  rollers  above  the  bearing,  or  by  means  of  pin-connected 
rocker  arms.  }  One  bearing  of  each  truss  should  be  fixed  against 
horizontal  movement,  in  order  to  resist  longitudinal  forces. 
(An  exception  was  the  Niagara  Railway  Suspension  Bridge, 
where  an  automatic  wedge  device,  for  dividing  the  expansion 
equally  between  the  two  ends  of  the  span,  was  provided.) 

19.  Braced-Chain  Construction. — Overhead  stiffening  trusses 
may  be  regarded  as  inverted  arches.  A  common  form  is  the 
three-hinged  truss  with  horizontal  lower  chord  (Type  3BH, 
Fig.  37),  and  these  are  designed  similar  to  spandrel-braced 
arches.  The  chords  and  web  members  are  built  up  of  plates 
and  angles  with  riveted  panel  points.  The  center  hinge  is 
designed  to  transmit  the  full  value  of  H  for  dead  and  live  load, 
and  the  maximum  vertical  shear  from  live  load.  At  the  towers, 
both  chords  are  supported  on  expansion  plates;  the  bottom 
chord  ends  there,  but  the  top  chord  passes  over  cast-steel  saddles 
and  continues  toward  the  anchorage.  The  top  chord  is  sup- 
ported at  the  top  of  the  towers  either  on  rockers  or  on  rollers 

(Fig-  37)- 

In  the  three-hinged  type  (3BH),  the  center  hinge  may  be 
located  either  in  the  upper  or  in  the  lower  chord.  In  the  former 
case,  the  upper  chord  will  carry  all  of  the  dead  load  and  full- 
span  live  load;  the  lower  chord  and  web  members  will  be 


104 


DETAILS  OF  CONSTRUCTION 


QJ  JL 


BRACED-CHAIN  CONSTRUCTION  105 

stressed  only  by  partial  loading.  Accordingly,  the  upper 
chord  will  have  a  fairly  uniform  section,  while  the  other  members 
will  be  comparatively  light.  This  arrangement  has  a  disad- 
vantage, however,  in  the  necessary  break  in  the  floor  system 
under  the  hinge;  the  stringers  must  be  provided  with  expansion 
joints,  and  the  wind  bracing  must  be  interrupted. 

If  the  hinge  is  placed  in  the  lower  chord,  false  members  are 
required,  for  the  sake  of  appearance,  to  cover  the  interruption 
in  the  top  chord.  Furthermore,  there  results  a  large  variation 
in  the  chord  stresses:  near  mid-span,  the  top  chord  stresses 
become  light  and  the  bottom  chord  stresses  become  heavy; 
and  the  reverse  occurs  near  the  towers. 

The  hinge  may  be  either  of  the  ordinary  pin  type  or  of  the 
plate  type.  In  the  latter  case,  the  chord  section  is  concentrated 
into  wide  horizontal  plates  to  connect  the  two  halves  of  the  span; 
and  the  vertical  shears  at  the  point  are  transmitted  by  means  of 
vertical  spring  plates. 

The  false  members  in  the  interrupted  chord  may  be  con- 
nected with  friction  bolts  in  slotted  holes,  so  that  the  resulting 
friction  may  act  as  a  brake  against  oscillations. 

To  eliminate  bending  moments  in  the  stiffening  truss  at  the 
tower,  the  end  member  of  the  lower  chord  may  be  suspended 
from  the  saddle;  or  it  may  simply  rest  on  an  expansion  bearing, 
at  the  tower,  with  the  end  vertical  omitted  from  the  truss. 
Another  arrangement  is  to  make  the  tower  integral  with  the 
main  span  truss,  the  tower  being  pivoted  at  the  base,  and 
the  side  span  having  only  a  hinged  connection  at  the  top  of 
the  tower. 

The  use  of  a  wire  cable  for  the  top  chord  had  an  illustration 
in  the  Lambeth  Bridge,  London;  but  the  details  did  not  con- 
stitute an  example  worth  copying.  For  long  spans  it  is  generally 
considered  that  the  overhead  bracing  system  cannot  compete 
with  the  suspended  stiffening  truss,  unless  a  wire  cable  is  used. 
In  any  case,  the  use  of  a  cable  as  a  truss  chord  gives  rise  to  diffi- 
culties in  the  detailing  of  connections  at  the  panel  points;  and, 
despite  noteworthy  studies  and  designs  (e.g.,  Fig.  39),  the  prob- 
lems involved  cannot  be  considered  as  fully  solved. 


106 


DETAILS  OF  CONSTRUCTION 


BRACED-CHAIN  CONSTRUCTION 


107 


108  DETAILS  OF  CONSTRUCTION 

When  the  braced-chain  (01  braced-cable)  system  is  used,  the 
web  members  should  preferably  not  be  connected  until  full  dead 
load  and  half  live  load  are  on  the  structure  at  mean  temperature. 
There  will  thus  be  accomplished  a  reduction  in  extreme  stresses 
in  the  web  members  and  lower  chords;  the  maximum  tension 
will  be  equal  to  the  maximum  compression  in  each  member,  and 
this  stress  will  be  only  the  arithmetic  mean  of  the  extreme 
stresses  that  would  be  produced  without  this  precaution.  How- 
ever, if  the  design  specifications  prescribe  stringent  allowances 
for  alternating  stresses,  the  reduction  in  sections  by  this  device 
will  not  be  material. 

The  truss  depth  at  the  crown  (Type  3£)  should  preferably 
be  between  0.15  and  0.3  of  the  sag  of  the  chain.  Sufficient 
depth  must  be  provided  to  take  care  of  the  shearing  stresses 
and  to  prevent  undue  flexibility  in  the  central  portion  of  the 
span;  but  excessive  depth,  besides  increasing  the  metal  required, 
impairs  the  desired  graceful  appearance  of  the  suspension  con- 
struction. If  the  hinge  is  omitted,  any  increase  in  crown-depth 
serves  to  augment  the  temperature  stresses. 

The  form  of  braced-chain  construction  (Type  2B  V)  pro- 
posed by  Linden  thai  for  his  first  Quebec  design  (Fig.  39),  and 
for  the  Manhattan  Bridge,  has  many  advantages.  The  bottom 
chord  is  bent  up  toward  the  towers,  so  as  to  obviate  the  necessity 
for  long  web  members.  The  requirement  of  extra  wind  chords 
at  the  ends  of  the  span  is  not,  comparatively,  an  important 
objection. 

Suspension  trusses  with  bracing  above  the  principal  chain 
(Type  3BL)  are  exemplified  in  the  Point  Bridge  at  Pittsburgh 
Fig.  38).  In  this  system  the  stiffening  chords  are  straight,  and 
the  bottom  chord  is  parabolic.  The  top  chord  members  have 
to  resist  both  tension  and  compression. 

A  system  having  many  advantages  is  the  Fidler  Truss 
(Type  SBC),  exemplified  in  Lindenthal's  second  Quebec  design 
(Fig.  40),  in  a  Tiber  bridge  at  Rome,  and  in  the  Tower  Bridge 
at  London.  In  this  system,  consisting  of  two  crescent-shaped 
half -arches,  both  chords  are  curved,  the  bottom  chord  having  a 
sharper  curvature  than  the  equilibrium  curve  for  full  load. 


110  DETAILS  OF  CONSTRUCTION 

The  parabolic  curve  passes  midway  between  the  two  chords,  so 
that  they  are  about  equally  stressed.  With  this  system  it  is 
possible  to  avoid  compressive  stresses  in  both  chords. 

In  the  foregoing  systems  (Types  3BL,  35C),  it  should  be 
noted  that  the  suspended  floor  system  must  be  interrupted,  or 
else  of  negligible  moment  of  inertia,  under  the  center  hinge.  It 
is  more  important  here  than  in  three-hinged  arches,  on  account 
of  the  greater  crown  deflections  in  suspension  systems. 

The  systems  using  parallel  chains  connected  with  web- 
bracing  have  been  little  used  on  account  of  the  difficulties  in 
stress  analysis.  If  each  chord  has  its  own  backstay  (Type  OBP) , 
the  system  is  threefold  statically  indeterminate.  If  the  top 
chord  is  interrupted  at  the  towers  (Type  2BP),  the  indetermi- 
nateness  is  reduced.  It  would  be  more  effective,  in  such  case,  to 
bring  the  two  chords  together  in  crescent  form  instead  of  using 
parallel  chords.  In  Lindenthal's  Seventh  St.  Bridge  at  Pitts- 
burgh and  in  his  first  design  for  the  North  River  Bridge,  the  bot- 
tom chord  rested  directly  on  bearings  on  the  top  of  the  towers, 
and  the  top  chord  was  connected  thereto  by  a  double  quad- 
rangular linkwork  equivalent  to  a  hinge;  this  made  the  system 
singly  indeterminate  (aside  from  the  redundancy  of  web  mem- 
bers). When  parallel  chains  are  used  in  the  side  spans,  the 
bottom  chord  may  be  connected  to  the  anchorage;  or  both 
chords  may  be  brought  together  at  a  common  pin  for  connec- 
tion to  the  anchor  chain. 

The  latest  example  of  parallel  chain  construction  is  Linden- 
thal's new  design  for  the  Hudson  River  Bridge  (Fig.  41).  The 
main  span  is  3240  feet,  and  the  bridge  is  continuous  at  the  towers 
(Type  OBP). 

20.  Wind  and  Sway  Bracing. — To  take  care  of  transverse 
wind  pressure  and  lateral  forces  from  moving  train  loads,  and  to 
carry  these  forces  to  the  piers,  systems  of  lateral  and  sway 
bracing  are  required. 

A  system  of  wind  bracing  must  be  provided  in  the  plane  of 
the  roadway,  since  the  principal  horizontal  forces  originate 
there.  Such  bracing  system  is  obtained  by  inserting  diagonals 
between  the  floorbeams,  so  as  to  form  a  horizontal  truss;  using 


WIND  AND  SWAY  BRACING 


111 


112  DETAILS  OF  CONSTRUCTION 

for  chords  either  the  bottom  chords  of  the  main  stiffening  truss 
(Figs.  26,  30,  32,  36,  37)  or  else  adding  extra  longitudinals  called 
wind  chords  (Figs.  38-41,  56). 

If  the  stiffening  truss  is  high  enough  to  afford  necessary 
clearances,  a  bracing  system  in  the  plane  of  the  top  chord  may 
be  added,  giving  a  closed  cross-section  to  the  structure  (Figs. 
26,  32,  36,  43). 

If  the  roadway  is  elevated,  vertical  sway  frames  of  cross- 
bracing  may  be  introduced  between  the  trusses  (Fig.  26) ;  or  the 
floorbeams  may  be  built  of  deep  latticed  construction.  In  such 
case,  a  single  plane  of  horizontal  wind  bracing  will  suffice. 

A  novel  method  of  transverse  bracing  is  introduced  in  the 
design  for  the  Hudson  River  Bridge  (Fig.  41)  in  the  form  of 
deep  floorbeams  (32  feet  high)  of  quadrangular  construction 
(Vierendeel  Girders);  the  rectangular  openings  are  used  as 
passageways  for  the  railway  tracks. 

Bearings  must  be  provided  at  the  towers  to  take  the  hori- 
zontal reactions  of  the  wind  truss  without  hindering  longitudinal 
expansion.  Vertical  pins  bearing  in  slotted  guides  may  be  used 
for  this  purpose. 

The  cables  or  chains  present  so  small  a  surface  to  the  wind 
as  to  require  no  wind  bracing;  or  at  most  they  may  be  con- 
nected together  by  horizontal  ties. 

On  account  of  the  inherent  stable  equilibrium  of  the  suspen- 
sion system,  the  wind  bracing  is  to  a  certain  extent  relieved  of 
its  duty. 

Braced-chain  systems  are  provided  with  a  single  wind  truss 
below  the  roadway,  using  either  the  lower  chords  of  the  main 
truss  (Type  3BH,  Fig.  37)  or  special  wind  chords  (Types  2BVj 
SBC,  etc.,  Figs.  38,  39,  40,  41).  In  addition,  transverse  sway 
frames  are  located  at  intervals  between  the  two  suspension 
trusses  in  the  planes  of  verticals  or  diagonals,  so  far  as  clearance 
requirements  permit  (Fig.  38).  In  comparison  with  other  types 
of  bridges,  but  little  material  is  needed  for  this  sway  bracing 
since,  in  the  first  place,  the  low  position  of  the  center  of  gravity 
makes  the  suspension  truss  stable  without  bracing,  and,  in  the 
second  place,  there  are  no  top  chord  compression  members  (in 


TOWERS  113 

most  of  these  types)  to  be  braced  against  buckling.  It  is  essen- 
tial, however,  to  provide  properly  designed  rigid  portal  and 
sway  bracing  between  the  legs  of  the  main  towers  (Figs.  37,  38). 

A  profusion  of  overhead  bracing,  besides  being  structurally 
unnecessary,  will  impair  the  graceful  appearance  sought  in 
suspension  constructions. 

A  center  hinge  in  the  stiffening  truss  introduces  complica- 
tions in  the  design  of  the  wind-bracing  system.  If  the  hinge  is, 
as  usual,  in  the  top  chord,  the  wind  bracing  must  follow  the  two 
central  diagonals  to  make  connection  at  the  hinge;  these  central 
diagonals  then  act  as  wind-chord  members,  and  their  sections 
must  be  increased  accordingly.  If  the  top  chord  hinge  lies 
above  the  roadway,  the  cross-bracing  in  the  two  central  panels 
connecting  with  the  hinge  has  to  be  omitted.  This  produces  a 
point  of  weakness  in  the  horizontal  bracing  system,  and  should 
be  avoided  in  long  spans  either  by  omitting  the  hinge  or  by 
locating  the  hinge  in  the  bottom  chord. 

Early  suspension  bridges  that  were  found  to  have  excessive 
lateral  deflections  and  oscillations  were  stiffened  by  means  of 
wind  cables  (wire  ropes)  placed  in  a  horizontal  plane  under  the 
roadway;  these  ropes  were  connected  to  the  floorbeams  and 
were  anchored  to  the  piers,  so  as  to  form  a  horizontal  suspension 
system  of  cable  and  stiffening  truss.  To  take  care  of  wind  in 
both  directions,  a  double  system  of  wind  cables  must  be  used, 
and  their  sag-ratio  should  be  made  as  large  as  possible  (Fig.  38). 
For  greater  stiffness,  straight  auxiliary  cables  have  sometimes 
been  added. 

The  efficacy  of  the  above-described  system  of  wind  cables  is 
doubtful,  since  it  is  ordinarily  impossible  to  give  the  ropes 
proper  initial  tension;  consequently  they  do  not  commence  to 
take  stress  until  the  horizontal  deflection  of  the  structure  exceeds 
a  certain  amount.  For  this  reason,  wind  cables  have  not  been 
relied  upon  for  modern  designs,  but  rigid  wind  trusses  have  been 
adopted  instead,  to  take  care  of  the  wind  pressures. 

21.  Towers. — For  purposes  of  discussion,  the  tower  may  be 
considered  as  composed  of  two  parts:  the  substructure  or  pier; 
and  the  tower  proper,  extending  above  the  roadway  and  sup- 


114  DETAILS  OF  CONSTRUCTION 

porting  the  cables  or  chains.  The  pier  does  not  involve  any 
special  features  differentiating  it  from  ordinary  bridge  piers. 

The  tower  must  be  designed  so  as  not  to  obstruct  the  road- 
ways. It  is  therefore  composed  of  a  column  or  tower  leg  for 
each  suspension  system  (Figs.  30,  35-38).  For  lateral  stability, 
the  tower  legs  are  braced  together  by  means  of  cross-girders  and 
cross-bracing  (Figs.  30,  35)  or  by  arched  portals  (Figs.  37,  38). 
The  sway  and  portal  bracing  are  necessary  to  brace  the  tower 
columns  against  buckling,  to  take  care  of  lateral  components 
from  cradled  cables  or  chains,  and  to  carry  wind  stresses  down  to 
the  piers. 

The  design  of  the  tower  depends  upon  the  material  employed. 
This  is  either  masonry  (Figs.  25,  27,  29,  33)  or,  more  commonly 
steel  (Figs.  15,  17,  28,  30-32,  34-41)  ,  and  occasionally  timber. 
If  masonry  is  used,  the  tower  may  consist  of  shafts  springing 
from  a  common  base  beneath  the  roadway  and  connected  together 
at  the  top  with  gothic  arches  (Fig.  25) ;  or,  for  smaller  spans,  the 
tower  may  consist  of  two  separate  tapering  shafts  or  obelisks. 

To  meet  architectural  requirements  and  to  express  resistance 
to  transverse  forces,  the  outline  of  the  tower  should  taper  toward 
the  top.  This  also  conforms  to  structural  requirements. 

The  tower  legs  must  be  designed  as  columns  to  withstand 
the  vertical  reaction  of  the  cables;  also  as  cantilevers  to  resist 
the  unbalanced  horizontal  tension.  The  latter  will  depend  upon 
the  saddle  design  (fixed  or  movable),  the  temperature  and  load- 
ing conditions,  and  the  difference  in  inclination  of  main  and 
side-span  cable  tangents  at  the  saddle.  Forces  due  to  wind 
pressure  on  the  cables,  towers,  and  trusses  must  also  be  pro- 
vided for. 

The  application  of  steel  to  suspension-bridge  towers  offers 
many  advantages.  The  lower  cost  permits  a  greater  height  in 
order  to  secure  a  more  favorable  sag-ratio.  The  thermal  expan- 
sion of  the  steel  tower  balances  that  of  the  suspenders,  so  as  to 
eliminate  serious  temperature  stresses  which  would  otherwise 
arise  in  indeterminate  types  (2F,  25,  OF,  OS). 

Steel  tower  columns  (Figs.  15,  17,  30,  36-39)  are  made  up  of 
plates  and  angles  to  form  either  open  or  closed  cross-sections; 


SADDLES  AND   KNUCKLES  115 

the  sides  may  be  either  latticed  or  closed  with  cover  plates. 
The  relative  dimensions  are  governed  by  the  usual  specifica- 
tions for  the  design  of  compression  members,  particularly  in 
respect  to  limiting  unsupported  widths  of  web  plates.  The 
cross-section  enlarges  toward  the  base,  or  outside  stiffening 
webs  are  added;  and  the  base  must  be  anchored  to  resist  the 
horizontal  forces  (Fig.  37). 

For  high  towers,  the  individual  legs  may  be  made  of  braced- 
tower  construction,  each  leg  consisting  of  four  columns  spread- 
ing apart  toward  the  base  and  connected  with  lacing  or  cross- 
bracing  (Fig.  31). 

Rocker  towers,  pin-bearing  at  the  base,  afford  the  most 
economical  and  scientific  design  for  bridges  of  longer  span. 
They  eliminate  the  stresses  from  unbalanced  horizontal  forces 
without  requiring  movable  saddle  construction.  The  most 
notable  examples  actually  constructed  are  the  Elizabeth  Bridge 
at  Budapest  (Fig.  34)  and  the  Cologne  Bridge  (Fig.  17).  (See 
also  Figs.  15,  39,  40.) 

If  rocker  towers  are  adopted,  they  must  be  secured  against 
overturning  during  erection.  This  may  be  accomplished  by 
temporary  connections  to  the  adjoining  truss  structure,  by 
wedging  or  bracing  at  the  base,  or  by  guying  the  upper  portions 
of  the  tower. 

For  foot  bridges  and  bridges  of  small  span,  the  simplest 
tower  construction  employs  a  stiffened  plate  for  each  leg,  the 
two  legs  being  braced  together  and  carrying  steel  castings  at 
the  top  to  hold  the  cables. 

If  timber  is  used,  each  cable  support  consists  of  four  battered 
posts  with  framed  bracing,  the  two  legs  thus  formed  being  con- 
nected at  the  top  with  cross-bracing. 

22.  Saddles  and  Knuckles. — The  cables  are  generally  con- 
tinuous over  saddles  on  top  of  the  towers. 

Designs  have  been  made  with  the  main  cables  terminating 
at  the  towers,  with  a  special  connection  at  the  top  of  the  towers 
to  the  backstay  cables  (e.g.,  Morison's  North  River  design). 
The  advantages  claimed  were  shorter  cable  strands  to  handle  in 
erection,  elimination  of  stresses  due  to  bending  of  the  cable 


116  DETAILS  OF  CONSTRUCTION 

over  the  towers,  and  the  possibility  of  increasing  the  section  of 
the  backstays  to  permit  steeper  inclination.  The  latter  advan- 
tages, however,  can  be  secured  with  continuous  cables  by  employ- 
ing certain  design  features. 

If  the  stress  in  the  backstay,  as  a  result  of  steeper  inclination, 
is  much  greater  than  in  the  main  cables,  auxiliary  strands  may 
be  incorporated  in  the  backstay  to  increase  its  section;  and 
provision  should  be  made  for  the  connection  of  these  auxiliary 
strands  to  the  saddle.  (An  example  is  the  Rondout  Bridge  at 
Kingston,  N.  Y.,  Fig.  56.) 

Cable  saddles  are  generally  made  of  cast  steel  (Fig.  32). 
They  are  either  bolted  to  the  top  of  the  tower  (Fig.  36)  or  pro- 
vided with  rollers  (Figs.  32,  33,  37). 

Where  fixed  saddles  are  used  (Fig.  36),  the  resultant  unbal- 
anced horizontal  forces  must  be  calculated  and  provided  for  in 
the  design  of  the  tower,  unless  the  tower  is  made  of  rocker  type 
(Figs.  15,  17, 34,  39,  40).  If  the  saddles  are  movable,  the  eccen- 
tricity of  the  vertical  reaction  under  various  loading  conditions 
must  be  provided  for. 

The  simplest  but  least  satisfactory  construction,  used  in 
some  smaller  bridges,  consists  of  fixed  saddles  over  which  the 
chain  or  cable  is  permitted  to  slide.  In  early  cable  bridges,  the 
wrapping  was  discontinued  near  the  towers,  and  the  wires  were 
spread  out  to  a  flat  section  to  pass  over  the  saddle  casting; 
this  is  objectionable  as  it  gives  access  to  moisture.  It  is  prefer- 
able to  give  the  saddle  a  cross-section  conforming  to  the  cable 
section;  to  reduce  wear  from  the  rubbing,  the  cable  may  be 
protected  by  a  lead  sleeve.  On  account  of  the  friction,  this 
arrangement  does  not  eliminate  the  unbalanced  horizontal  pull 
on  the  top  of  the  tower.  On  the  whole,  this  construction,  or 
any  sliding  saddle  arrangement,  is  not  to  be  recommended. 

Another  saddle  arrangement  consists  of  steel  pulleys,  free  to 
rotate,  over  which  the  cable  passes  (Fig.  27).  A  similar  arrange- 
ment used  with  chains  consists  of  a  fixed  roller  nest  over  which 
curved  eyebars  slide  (Fig.  33) ;  the  resulting  bending  stresses, 
however,  are  objectionable. 

The  best  arrangement  to  permit  horizontal  movement  on 


SADDLES  AND  KNUCKLES  117 

top  of  the  tower  consists  of  a  roller  support  for  the  saddle  (Figs. 
32,37).  In  modern  designs,  the  rollers  are  of  equal  height  between 
two  plane  surfaces.  The  construction  comprises  a  bed  plate,  a 
nest  of  rollers  connected  by  distance  bars,  and  the  superimposed 
saddle  casting.  In  the  saddle  rests  the  cable  (Fig.  32)  which  is 
held  from  sliding  by  friction  or  clamping.  Instead  of  a  saddle, 
the  movable  part  may  consist  of  a  casting  to  which  the  chains 
are  pin-connected  (Fig.  37).  The  resultant  of  the  tensions  of 
the  cables  or  chains  should  pass  through  the  middle  of  the 
roller  nest  to  give  an  even  distribution  of  stress.  The  friction 
of  the  rollers  is  so  small  that  the  obliquity  of  the  resultant 
reaction  is  negligible. 

Instead  of  circular  rollers,  segmental  rollers  (rockers)  may 
be  used,  so  as  to  furnish  a  greater  diameter  and  thereby  reduce 
friction  and  roller-bearing  stress.  Segmental  rollers,  however, 
must  be  secured  against  excessive  motion  liable  to  cause  over- 
turning. 

Rollers  (Figs.  32,  33,  37)  serve  to  reduce  the  bending  stresses 
on  the  towers  due  to  unbalanced  horizontal  cable  pull  resulting 
from  temperature  and  special  loading  conditions.  On  the  other 
hand,  they  add  expense,  increase  erection  complications,  give 
trouble  in  maintenance,  and  merely  substitute  eccentric  vertical 
loading  for  unbalanced  horizontal  pull.  On  the  whole,  fixed 
saddles  provide  a  simpler  and  safer  construction. 

Another  saddle  design  consists  of  a  rocker,  pin-connected  at 
either  upper  or  lower  end  to  the  tower  and  carrying  the  cable  or 
chain  at  the  other  end.  The  rocker-hanger  or  pendulum  type 
has  been  used  only  in  earlier  bridges ;  the  main  objection  to  it  is 
the  increased  height  of  tower  required.  The  rocker-post  type 
(Fig.  39)  has  pin-connection  to  cable  or  chain  at  the  upper  end, 
and  has  a  pin  or  cylindrical  bearing  at  the  lower  end. 

Short  rocker  posts  should  not  be  used  for  long  spans;  after 
such  posts  assume  an  inclined  position  under  temperature  varia- 
tion, the  return  to  normal  position  is  seriously  resisted  by  the 
necessity  of  raising  the  point  of  cable  support  through  a  vertical 
height. 

The  tower  itself  may  be  made  to  serve  as  a  rocker  post  for  its 


118  DETAILS  OF  CONSTRUCTION 

full  height  by  providing  hinge  action  at  the  base.  This  construc- 
tion was  used  in  1857  for  a  bridge  over  the  Aare  at  Berne;  also 
in  the  Elizabeth  Bridge  at  Budapest,  1903  (Fig.  34),  in  Linden- 
thal's  designs  for  the  Quebec  Bridge,  1899,  1910  (Figs.  39,  40) 
and  for  the  Manhattan  Bridge,  1902;  in  the  Rhine  Bridge  at 
Cologne,  1915  (Fig.  17),  and  in  the  design  for  the  Detroit  Bridge, 
1921  (Fig.  15).  Instead  of  using  pins  at  the  lower  end,  the 
hinge-action  may  be  secured  by  providing  the  tower  leg  with 
a  segmental  base  (Fig.  17)  or  with  a  concave  nest  of  rollers 

(Fig.  15). 

Knuckles  are  provided  in  the  anchorages  at  all  points  where 
the  backstays  or  anchor  chains  alter  their  direction  (Figs.  29,  32, 
33,  36-40,  42).  They  are  similar  in  function  to  tower  saddles, 
and  should  be  designed  to  permit  any  movement  due  to  thermal 
or  elastic  elongation  of  the  anchor  chain. 

Sliding  bearings  (Figs.  37,  39)  are  commonly  used  at  knuckle 
joints,  and  may  be  considered  suitable  where  the  directional 
change  is  small.  In  the  Rondout  Bridge  at  Kingston,  N.  Y. 
(Fig.  56),  the  design  consists  of  vertical  pin  plates  supporting  the 
knuckle  pin  and  sliding  on  steel  plates  anchored  in  the  masonry. 

Roller  bearings  (Figs.  33,  36,  40)  give  a  better  design  for  the 
anchorage  knuckles.  A  cable  may  be  carried  in  a  saddle  casting 
resting  on  rollers  (Fig.  36) ;  and  chain  eyebars  may  be  either 
directly  supported  on  rollers  (Fig.  33)  or  may  be  pin-connected 
to  a  casting  carried  on  rollers  (Fig.  40).  The  plane  of  the 
rollers  should  be  normal  to  the  bisector  of  the  angle  of  the  chain 
or  cable. 

Rocker  supports  (Figs.  32, 39, 42)  are  also  used  for  anchorage 
knuckles.  The  change  in  direction  may  be  accomplished  at 
one  main  rocker  strut  (Fig.  42),  or  may  be  distributed  over  a 
large  number  of  small  rocker  knuckles  (Fig.  32).  The  direction 
of  the  rocker  should  preferably  coincide  with  the  bisector  of  the 
angle  of  the  chain  or  cable. 

23.  Anchorages. — The  safety  of  a  suspension  bridge  depends 
upon  the  security  of  the  anchorages.  Consequently,  in  any 
new  design,  the  anchorages  should  receive  thorough  study  and 
their  construction  should  be  carefully  supervised;  and,  after 


ANCHORAGES 


119 


120  DETAILS  OF  CONSTRUCTION 

completion,  the  condition  of  the"  anchorage  should  receive 
watchful  attention.  Accessibility  for  inspection  and  main- 
tenance should  be  considered  in  the  design. 

In  rare  cases  it  is  possible  to  anchor  the  cables  in  natural 
rock  (Figs.  27,  39,  40).  The  shaft  or  tunnel  which  is  to  contain 
the  anchor  chain  must  then  be  driven  to  such  depth  as  to  reach 
and  penetrate  rock  that  is  perfectly  sound,  proof  against  weather- 
ing and  of  sufficient  thickness  to  afford  the  necessary  anchorage. 

In  most  cases,  the  anchorage  requires  a  masonry  construction 
which  resists  the  cable  pull  by  friction  on  its  base  or  by  the  resist- 
ing pressure  of  the  abutting  earth  (Fig.  34). 

Masonry  anchorages  may  be  imbedded  below  ground  level, 
with  backstays  connecting  them  to  the  nearest  towers  (Fig.  29) ; 
or  they  may  constitute  the  end  abutments  of  the  side  spans 
(Figs.  26,  28,  30,  32-38,  41).  The  latter  arrangement  gener- 
ally requires  bending  or  curving  the  line  of  the  cable  or  chain 
from  its  initial  inclination  to  a  more  vertical  direction,  in  order 
to  secure  the  necessary  depth  of  anchor  plate  without  excessive 
length  of  anchorage  (Figs.  32,  33,  36-38,  42).  In  addition  to 
stability  against  sliding,  such  anchorage  must  also  be  designed 
for  stability  against  tilting  or  overturning.  Furthermore,  the 
applied  forces  must  be  followed  through  the  masonry  and  the 
resulting  normal  and  shearing  stresses  at  all  sections  and  joints 
must  be  provided  for.  The  extreme  pressures  on  the  base  should 
also  be  investigated,  to  make  sure  that  they  do  not  exceed  the 
allowable  load  on  the  foundation  material  (Fig.  44).  Founda- 
tions on  piles  (Figs.  32,  36)  should  be  avoided,  as  they  give 
insufficient  security  against  displacement  of  the  anchorage;  if 
such  foundations  are  unavoidable,  an  ample  proportion  of  batter 
piles  should  be  provided. 

The  anchor  chains  go  through  the  masonry  and  are  fastened 
at  their  ends  to  anchor  plates  or  to  reaction  girders.  In  larger 
structures,  anchor  tunnels  may  be  left  in  the  masonry,  affording 
access  for  inspection  of  the  anchor  chains  (Figs.  33,  34,  37,  42). 

As  a  rule,  each  cable  or  chain  is  separately  anchored;  in  rare 
cases,  the  two  cables  have  been  connected  in  the  anchorage  so  as 
to  form  a  loop  around  a  body  of  masonry  or  rock. 


ANCHORAGES  121 

The  anchor  chains  are  commonly  secured  by  means  of  an 
anchor  plate  which  is  either  a  single  casting  or  built  up  of  cast 
sections;  the  last  link  is  passed  through  this  anchor  plate  and 
is  fastened  behind  it  with  a  special  pin  or  bolt  (Figs.  33,  37,  38). 
The  anchor  plate  is  stiffened  against  bending  by  means  of  per- 
pendicular webs  or  ribs  (Figs.  33,  38);  it  must  [have  a  bearing 
surface  large  enough  to  transfer  and  distribute  the  pressure  to  a 
sufficient  area  and  mass  of  masonry. 

Instead  of  cast-steel  anchor  plates,  anchor  girders  built  up  of 
rolled  sections  have  been  used  in  more  recent  designs  (Figs.  32, 
39,  40,  42).  The  chains  are  pin-connected  to  the  webs  of  these 
girders,  and  the  latter  transmit  the  reaction  to  grillages  or  cast- 
ings bearing  against  the  masonry. 

Provision  for  adjustment  of  backstay  length  may  be  made  in 
the  anchorage,  if  not  elsewhere.  Adjustment  may  be  secured 
through  the  use  of  wedges  in  the  connection  behind  the  anchor 
plate  (Fig.  37);  or  by  means  of  a  threaded  connection  between 
the  strand  sockets  and  round  rods  passing  through  the  anchor 
plate. 

The  anchor  plates  (Figs.  33, 37, 38)  are  designed  like  any  other 
masonry  plates.  The  area  is  determined  by  the  allowable 
bearing  pressure  on  the  concrete  or  stone,  and  the  section  of  the 
plate  and  stiffening  ribs  are  determined  by  the  shearing  and 
bending  stresses.  The  holding  bolts  or  wedges  must  also  be 
proportioned  for  shearing  and  bending  stresses;  for  greater 
bending  strength,  the  bolts  may  be  made  of  oval  rather  than 
circular  section. 

A  connection  of  cable  to  anchor  chain  is  illustrated  in 
Fig.  32. 

In  small  bridges,  the  cable  is  often  anchored  directly,  by 
passing  the  wire  slings  around  anchored  bolts  or  around  anchor 
blocks.  This  method  can  be  used  only  with  parallel  wire  strands1, 
and,  if  the  diameter  of  the  sling  is  too  small,  excessive  bending 
stresses  will  arise  in  the  wire. 

At  each  change  in  direction  of  the  cable  or  chain  in  the 
anchorage,  a  bearing  is  required.  This  may  consist  of  a  casting 
with  rounded  surface  over  which  the  cable  or  chain  may  slide 


122  DETAILS  OF  CONSTRUCTION 

(Figs.  37,  39),  or  of  a  flat  plate  on  which  the  eyebar  heads  rest. 
A  knuckle  casting  with  bearing  for  the  eyebar  pin  is  preferable 
to  one  on  which  the  eyebar  heads  bear.  If  the  change  in  direction 
of  the  anchor  chain  is  considerable,  roller  bearings  (Figs.  33, 
36,40)  or  rockers  (Figs.  32,  42)  must  be  provided. 

In  general,  aside  from  greater  simplicity,  a  straight  anchor 
chain  (Figs.  30,  34)  is  preferable  to  a  curved  or  bent  chain  (Figs. 
29, 32, 33, 36-38,  42),  as  the  latter  results  in  greater  lengthening  of 
the  cable  from  compression  or  settlement  of  the  masonry.  Space 
limitations,  however,  frequently  make  this  arrangement  unavoid- 
able. 

If  it  is  desired  to  leave  the  anchorage  steel  accessible,  shafts 
or  tunnels  must  be  provided  in  the  masonry,  large  enough  for  a 
man  to  pass  through  (Figs.  27,  29,  33,  34,  37,  42);  a  clearance  of 
2  to  3  feet  is  necessary.  Near  the  lower  end,  the  shaft  generally 
becomes  constricted  in  order  to  reduce  the  required  size  of  the 
anchor  plate;  consequently  the  end  of  the  chain  is  not  fully 
accessible  to  inspection.  For  the  examination  of  the  anchor 
plate  and  fastenings,  vaulted  chambers  or  horizontal  passage- 
ways (about  3  to  4  feet  wide  and  5  to  6  feet  high)  are  provided 
behind  the  anchor  plate  (Figs.  33,  37,  38,  42).  These  chambers 
may  lead  to  the  sides  of  the  anchorage,  where  they  are  sealed  by 
doors  (Figs.  33,  42),  or  they  may  be  reached  through  horizontal 
tunnels  (Fig.  27).  Inclined  shafts  (Figs.  27,  33,  37,  42)  may  be 
roofed  with  stepped  slabs,  flat  slabs  or  an  arched  vault.  Rain 
and  dirt  must  be  excluded,  and  the  points  of  emergence  of  the 
cables  or  chains  should  be  protected  accordingly. 

Instead  of  leaving  open  passageways  for  inspection  and 
painting,  the  opposite  course  may  be  adopted  and  the  anchor- 
age completely  sealed  against  the  entrance  of  air  or  water  (Figs. 
39,  40,  42).  In  such  designs,  the  anchorage  steel  is  imbedded 
in  concrete,  or  surrounded  with  waterproofing  material,  so  as  to 
exclude  air  and  moisture  and  thereby  prevent  oxidation.  The 
shafts  receiving  the  anchor  chains  or  cables  are  made  as  narrow 
as  possible,  and  are  subsequently  filled  with  concrete,  cement 
mortar,  asphaltic  cement  or  other  waterproofing  substance. 

The  anchorage  may  be  built  of  stone  masonry  or  of  concrete. 


ANCHORAGES  123 

The  use  of  reinforced  concrete  gives  maximum  flexibility  in 
design. 

The  forces  acting  on  the  anchorage  as  a  whole  are  the  cable 
pull,  the  weight  of  the  masonry  and  any  superimposed  reactions, 
and  the  pressure  or  resistance  of  abutting  earth.  For  the  study 
of  the  internal  stresses,  the  outside  cable  pull  is  replaced  by  the 
reactions  of  the  anchor  plates  and  the  knuckle  castings.  By 
graphic  composition  of  these  various  applied  forces,  the  lines  of 
pressure  in  the  masonry  are  determined.  The  resultant  of  all 
the  external  forces,  including  the  weight  of  the  anchorage,  must 
intersect  the  base  within  the  limits  necessary  to  prevent  uplift 
at  the  heel  (Fig.  44) ;  and  the  inclination  of  the  resultant  from 
the  normal  must  not  exceed  the  angle  of  friction.  If  it  proves 
impracticable  to  secure  this  stability  against  sliding  with  a  hori- 
zontal foundation,  the  base  may  be  sloped  (Fig.  42)  or  stepped 
to  increase  the  sliding  resistance.  Stepping  the  base  is  not 
effective  save  on  hard  foundations.  In  soft  ground,  requiring 
piles,  the  pile  caps  should  be  imbedded  in  masonry;  and  the 
piles  should  preferably  be  battered  in  the  direction  of  the 
resultant  pressure. 

Granite  or  other  stone  blocks  should  preferably  be  used  to 
take  the  direct  pressure  of  the  anchor  plate  and  knuckle  cast- 
ings (Figs.  32,  33,  37,  42).  Extending  forward  from  the  anchor 
plate,  cut-stone  blocks  may  be  laid  in  arch  formation,  following 
the  curving  line  of  resultant  pressure  and  with  joints  normal 
thereto.  The  rest  of  the  anchorage,  serving  only  to  provide 
weight,  may  be  built  of  rubble  masonry  or  brickwork  in  hori- 
zontal courses;  or  of  rubble  concrete. 

Special  designs  of  anchorages  are  frequently  necessitated  by 
physical  conditions  and  economic  limitations.  In  one  design 
for  a  South  American  bridge,  the  author  devised  a  buttressed 
concrete  box  filled  with  sand  to  give  it  weight.  In  another 
design,  he  used  a  concrete  tower  encasing  the  anchor  chain  and 
supporting  its  saddle,  the  tower  being  braced  by  a  flying  but- 
tress of  concrete  delivering  the  resultant  to  an  inclined  founda- 
tion. For  the  design  of  the  Detroit- Windsor  Bridge  (Fig.  15), 
on  account  of  the  depth  to  rock,  there  was  devised  by  C.  E. 


124  DETAILS  OF  CONSTRUCTION 

Fowler  and  the  author  an  articulated  type  of  anchorage  con- 
sisting of  two  members:  an  anchor  chain  in  an  inclined  shaft 
leading  into  rock,  and  a  heavy  steel  strut  in  an  oppositely 
inclined  shaft  carrying  the  resultant  thrust  to  rock,  both  shafts 
being  rilled  with  waterproof  concrete  for  the  protection  of  the 
steel;  the  two  members  form  an  inverted  V,  the  cable  being 
attached  at  their  junction. 


CHAPTER  III 
TYPICAL  DESIGN  COMPUTATIONS 

(NOTE.— All  references  are  to  Chapter  I,  "Stresses  in  Suspension  Bridges") 


EXAMPLE  1 

Calculations  for  Two-hinged  Suspension  Bridge  with 
Straight  Backstays  (Type  2F) 

1.  Dimensions. — The  following  dimensions  are  given: 

/  =  Main  Span  =  50  panels  at  22.5  ft.  =  1125  ft.     (/'  =  /). 

/2  =  Side  Span  =281. 25  ft.  =-. 

4 

/  =  Cable-sag  in  main  span  =  112. 5  ft.     \n=j=  —  )• 

/i  =  Cable-sag  in  side  spans  =o.  (Straight  backstays). 
d  =  Depth  of  Truss  =  22. 5  ft. 

Mean  Chord  Sections  (gross) : 

Top  =  94  sq.  in.     Bottom  =161  sq.  in. 

7  =  Mean  Moment  of  Inertia  of  stiffening   truss   in  main 
span  =  94(14.2)2+ i6i(8.3)2  =  30,000  in.2  ft.2  (i  truss). 
Width,  center  to  center  of  trusses  or  cables  =34.5  ft. 

A  =  Cable  Section  =  84  sq.  in  per  cable.     (A  i  =  A) . 
tan  a  =  Slope  of  Cable  Chord  in  main  span  =  o. 

tan  «i  =  Slope  of  Cable  Chord  in  side  span  =4^=4^=0.4. 

2.  Stresses  in  Cable. — Given: 

w  =  Dead  Load  per  cable  (including  cable) 
=  2650  Ib.  per  lin.  ft. 

125 


126  TYPICAL  DESIGN   COMPUTATIONS 

^'=Live  Load  per  cable  =  850  Ib.  per  lin.  ft. 
/  =  Assumed  Temperature  Variation  =  ±60°  F. 
(£w/  =  i  i,  7  20  Ib.  per  sq.  in.) 

All  values  are  given  per  cable. 

For  Dead  Load,  by  Eq.  (n),  the  horizontal  component  of 
cable  stress  is, 

wl2     10   ,  ,  .          f    ,  .  ,,  . 

kips,     (i  kip  =  1000  Ib.) 


q/ 

For  Live  Load,  by  Eq.  (167),  the  denominator  of  the  formula 
for  H  is, 


By  Eq.  (168),  the  live-load  tension  will  be, 

p'l       2 
H=~^n  =  ~]S['l)'l=  I'I46  P'l=  I095  kips. 

The  total  length  of  cable  between  anchorages  is  given  by  Eq. 

(176): 

=(i+f»2)  +  2-  sec  ai  =  (i  +  .o27)+i(i.o8)  =  1.567. 


Then,  for  temperature,  by  Eq.  (156), 

3E/co/L        3(30,000)  1.567 
H'=    '^NT  =T7T^F  T^(II' 

Adding  the  values  found  for  H: 

D.  L.         3730  kips 
L.  L.          1095 
Temp.  75 


we  obtain,  Total  H  =  4900  kips  per  cable. 

The  maximum  tension  in  the  cable  is,  by  Eq.  (5), 
Ti=H-sec  ai=H(i.o$)  =  5300  kips. 

At  65,000  Ib.  per  sq.  in.,  the  cable  section  required  is  5300 ^ 
=  82  sq.  in.     (Section  provided  =  84  sq.  in.) 


TWO-HINGED  BRIDGE  WITH  STRAIGHT  BACKSTAYS       127 


3.  Moments  in  Stiffening  Truss.  —  Given: 

Live  Load  =  p  =  1600  Ib.  per  lin.  ft.  per  truss. 
(All  values  given  and  calculated  are  per  truss.) 
With  the  main  span  completely  loaded,  the  bending  moment 
at  any  section  x  is  given  by  Eq.  (169)  : 


Total  Jf- 


-        -i#*(*-*)to85). 


In  other  words,  only  8.5  per  cent  of  the  full-span  live  load  is 
carried  by  the  truss.     Accordingly,  at  the  center, 

pi2 
Total  M  =  .085—  =  +21,500  ft.  kips  per  truss. 

o 

At  other  points,  the  values  of  M  are  proportional  to  the  ordinates 
of  a  parabola.     They  are  obtained  as  follow  s: 


Section  (-  j 

Parabolic  Coefficient 

Total  M  (ft.  kips) 

o 

4X     o  X     1=0 

o 

O.I 

4X    .1   X    .9  =    -36 

+  7,8oo 

O.2 

4X    .2  X    .8  =    .64 

+  13,800 

0-3 

4X    .3  X    .7  =    -84 

+18,100 

0.4 

4X    .4  X    .6  =    .96 

+  20,000 

0-45 

4X    -4SX    .55  =    -99 

+21,300 

o-5 

4X    .5  X    .5   =1.00 

+21,500 

For  maximum  and  minimum  moments,  the  critical  points 
are  found  by  solving  Eq.  (142)  : 


The  values  of  the  minimum  moments  are  then  given  by  Eq. 

(170): 


Min. 


=  -2px(l~X^D(k)  =  - 
=  -5.52(TotalM) 


(Total  M}  -D(k) 


The  tabulations  hi  Table  I  or  the  graphs  in  Fig.  1  2  are  used  in 
solving  the  values  of  C(k)  for  k  and  then  finding  the  correspond- 


128 


TYPICAL  DESIGN  COMPUTATIONS 


ing  values  of  D(k).    The  following  tabulation  shows  the  succes- 
sive steps: 


X 

1 

y 

i 

X 

y 

C(k) 

k 

D(k) 

Min.  M 

(ft.  kips) 

o 

o 

(2.50) 

(-436) 

(-355) 

(-531) 

o 

.1 

.036 

2.78 

.485 

•392 

•437 

-18,800 

.2 

.064 

3.12 

•544 

•437 

•334 

-25,400 

•3 

.084 

3-57 

.623 

•497 

.224 

—  22,400 

•  4 

.096 

4-17 

.728 

.584 

•113 

-12,800 

•45 

.099 

4-55 

•795 

-647 

.o6i+.ooi 

—  7,300 

•5 

.100 

5.00 

.872 

.729 

.024-}-  .024 

-  5,7oo 

•55 

.099 

5-55 

.970 

.876 

.001  +  .061 

-  7,300 

(By  using  Fig.  13,  D(k}  may  be  found  directly  from  the  values  of  C(k).) 


and  the  symmetrical 


N 

For  all  sections  between  x=  —  -I  —  . 

4 

point  a;  =  .5647,  a  correction  is  made  in  the  above  table  for  the 
second  critical  point,  as  explained  under  Eq.  (143  '). 
To  find  the  maximum  moments,  apply  Eq.  (144)  : 

Max.  M  =  Total  M-Min.  M. 

Dividing  the  maximum  and  minimum  moments  by  the  truss 
depth  (^  =  22.5  ft.),  we  obtain  the  respective  chord  stresses  as 
follows: 


Section 

Chord  Stresses  (kips) 

X 

Maximum  M 

I 

(ft.  kips) 

Maximum 

Minimum 

o 

o 

o 

0 

O.I 

+26,600 

=Fu8o 

=h  840 

0.2 

+39,200 

=Fi74o 

±1130 

0-3 

+40,500 

Ti8oo 

rtlOOO 

0.4 

+33,400 

^1490 

±    570 

0-45 

+  28,600 

^1270 

rb   320 

0-5 

+  27,200 

Tl2IO 

±    250 

TWO-HINGED  BRIDGE  WITH  STRAIGHT  BACKSTAYS       129 


In  this  tabulation  of  the  stresses,  the  upper  signs  refer  to  the  top 
chord  and  the  lower  signs  to  the  bottom  chord.  Dividing  the 
above  values  by  the  specified  unit  stresses  in  tension  and  com- 
pression, respectively,  we  obtain  the  required  net  and  gross 
sections  of  the  chord  members.  For  the  bottom  chords,  these 
sections  must  be  increased  to  provide  for  the  wind  stresses, 
computed  as  indicated  below.  In  addition,  the  temperature 
stresses  must  be  taken  into  consideration. 

The  moments  produced  by  temperature  variation  are  given 
by  Eq.  (157): 

Mt=-Ht.y. 

As  found  above,  #,=  1=75  kips  for  a  temperature  variation  of 
±60°  F.  At  the  center, 

Jf«=±(75  kipsXii2.5  ft.)  =  ±8450  ft.  kips. 

At  other  sections,  the  moments  are  proportional  to  the  parabolic 
ordinates  y : 


Section 

Parabolic 
Coefficient 

Mt 

X 

r=0 

0 

0 

.  i 

•36 

±3040 

,2 

.64 

±5400 

•3 

.84 

±7100 

•  4 

.96 

d=8lOO 

•  5 

I.OO 

±8450 

(The  upper  signs  pertain  to  a  rise  in  temperature,  the  lower  signs  to  a  fall.) 

The  resulting  stresses  are  to  be  combined  with  the  live-load 
stresses  previously  found,  in  whatever  manner  the  specifica- 

x 

tions  may  prescribe.  For  the  sections  y  =  o  to  0.4,  the  tempera- 
ture moments  amount  to  less  than  25  per  cent  of  the  live-load 
Max.  M,  in  which  case,  according  to  some  specifications,  the 
temperature  stresses  may  be  ignored. 


130  TYPICAL  DESIGN  COMPUTATIONS 

4.  Shears  in  Stiffening  Truss.— 

(p  =  i6oo  Ib.  per  lin.  ft.,  %pl  =  goo  kips.) 


With  the  main  span  fully  loaded,  the  shears  at  the  various 
sections  are  given  by  Eq.  (173): 


Total  F=- 


Section 

(-¥) 

Total  V 

X 

r° 

I 

76  kips 

.1 

.8 

61 

.2 

.6 

46 

•3 

•  4 

3i 

•  4 

.2 

15 

•5    . 

O 

o 

The  maximum  shears  are  given  by  Eq.  (149) 


The  values  of  G(T)  are  taken  from  Table  I,  and  the  shears  are 
obtained  as  follows: 


Section 

(-f)' 

i/£   ^ 

^2       */ 

•® 

[] 

Maximum  V 

X 

r° 

i 

2.2Q 

.400 

.084 

+  76+219 

.1 

.81 

1.84 

.482 

.114 

+  83+110 

.2 

.64 

1-38 

.565 

.220 

+  131+  28 

•3 

•49 

O.Q2 

.647 

.405 

+  179 

•4 

.36 

0.46 

.726 

.666 

+216 

•5 

•25 

O 

.800 

i  .000 

+  225  kips 

TWO-HINGED  BRIDGE  WITH  STRAIGHT  BACKSTAYS       131 

For  all  sections  x<-(i )=.282/,  the  loading  for  maximum 

shear  extends  from  the  given  section  x  to  a  critical  point  kl 
denned  by  Eq.  (150): 


4  /— 


.431. 

2X 


The  values  C(Jfc)  are  solved  for  k  with  the  aid  of  Fig.  13. 


2X 

Section 

'" 

C(k) 

k 

0 

I 

•  436 

•355 

.  i 

.8 

•545 

•  438 

.2 

.6 

•730 

.588 

For  these  sections,  a  correction  is  to  be  added  to  the  values  of 
Max.  V  found  above.     This  additional  shear  is  given  by   Eq. 


Add. 


Section 

£ 

•_Y«_£\ 

G(k) 

r   -i 

Add.  V 

7V\2       // 

L    J 

X 

•355 

.416 

2.29 

.691 

•585 

+2  19  kips 

.1 

•  438 

.316 

1.84 

•755 

•389 

+  110 

.2 

•  588 

.170 

1.38 

•  858 

.185 

+  28 

(By  using  Fig.  13,  G(k]  may  be  found  directly  from  the  values  of  C(k).) 

The  minimum  shears  are  then  given  by  Eq.  (153) : 
Min.  V  =  Total  F-Max.  V. 


132 


TYPICAL  DESIGN  COMPUTATIONS 


Section 

Total  V 

Maximum  V 

Minimum  V. 

X 

r° 

+76  kips 

+295  kips 

—  219  kips 

.1 

+61 

+  iQ3 

-132 

.2 

+46 

+  IS9 

-113 

•3 

+3i 

+  179 

-148 

•  4 

+  15 

+216 

—  2OI 

•  5 

o 

+225 

-225 

The  temperature  shears  are  given  by  Eq.  (158) : 

Vt=—Ht'(ten.  0-tana). 
In  this  case,  tan  a  =  o,  and  Ht==F'j5  kips.     At  the  ends, 

tan  0  =  y-  =  4n  =  0.40, 
and  the  slope  (tan  0)  diminishes  uniformly  toward  the  crown. 


Section 

tan  <f> 

vt 

X 

r° 

0.40 

±30  kips 

.1 

•32 

±24 

.2 

•24 

±18 

•3 

.16 

±12 

•4 

.08 

±  6 

•5 

o 

o 

These  shears  are  to  be  combined  with  the  maximum  and  mini- 
mum live-load  shears,  as  may  be  required  by  the  specifications. 

6.  Wind  Stresses  in  Bottom  Chords. — 

(Assumed  wind  load =p  =  400  Ib.  per  lin.  ft.) 

If  the  lateral  bracing  is  in  the  plane  of  the  bottom  chords, 
these  chords  act  as  the  chords  of  a  wind  truss.  The  applied 
wind  pressure  p  is  partly  counteracted  by  a  force  of  restitution  r 
due  to  the  horizontal  displacement  of  the  weight  of  the  stiffening 


TWO-HINGED  BRIDGE  WITH  STRAIGHT  BACKSTAYS       133 

truss  w.     The  resulting  reduction  in  the  effective  horizontal  load 
is  given  with  sufficient  accuracy  by  the  formula, 

wl* 
r       '°I 


p  wl*' 

- 


(For  the  derivation  of  this  formula,  see  Steinman's  "Suspension 
Bridges  and  Cantilevers,"  D.  Van  Nostrand  Co.,  1913,  page  76.) 
In  this  case,  w  =  total  dead  load  (both  trusses)  =  5300  Ib.  per  lin. 
ft.;  v  =  vertical  height  from  cable  chord  to  center  of  gravity  of 
the  dead  load  =  13  5  it.;  /  =  moment  of  inertia  of  wind  truss  = 
f(i6i)x(34.5)2  =  96,000  in.2ft.2  Substituting  these  values,  we 
obtain, 

r=    .284 

/>~i+.284~ 

Hence  the  force  of  restitution  r  (due  to  the  obliquity  of  suspen- 
sion after  horizontal  deflection)  amounts,  in  this  case,  to  22  per 
cent  of  the  applied  wind  load  (p)  at  the  center  of  the  span. 
The  force  r  diminishes  to  zero  at  the  ends  of  the  span,  and  the 
equivalent  uniform  value  of  r  may  be  taken  as  f  of  the  mid-span 
value.  The  resultant  horizontal  load  on  the  span  is, 

/>-fr  =  400-4(88)  =327  Ib.  per  lin.  ft. 

Treating  this  value  as  a  uniform  load,  the  bending  moment  at 
the  center  is, 

Mw=¥£-  =  5i,goo  ft.  kips. 

Dividing  by  the  truss  width  34.5  ft.,  we  obtain  the  chord  stress 
=  dz  1  500  kips  at  mid-span.  The  wind  stresses  at  other  sections 
will  be  proportional  to  parabolic  ordinates,  being  zero  at  the 
ends  of  the  span. 

The  shears  in  the  lateral  system  may  also  be  calculated  for 
the  resultant  uniform  load  of  327  Ib.  per  lin.  ft. 


Maximum  Horizontal  Component  =  3,870,000  Ibs. 
Maximum  Teusion  in  Cable  =  4,300,000  Ibs. 

Area  =C8  sq.  iu. 


Two  parallel  wire  cables  10  diameter  composed  of  seven  c 
each  containing  336  No.  6  E.G.  (.102  diameter)  galvanize 
Total  wire  in  two  cables  4704.  Cables  to  be  continuous  o\ 
Cables  to  be  wrapped  with  No.  9  gahauized  steel  wrappi 


Mainland 

Bteel  for  floor  over  anchorages 
4  Panels  @24'G" 


3  Floor  beams 

f2.Eleetric  24"ls  80  ib's. 
T  •  ,«  J  5-Roadway  15"ls42  Ibs. 
Joists  ^  j..^,,,.  pil)e  16"u  33  ll)B> 

^3-Sidewalk  12"i!i  25  Iba. 
Top  of  Roadway    +  2  $  Grade 


4  Ls  8  x  3  x 

2-Pls.  42  x  %  Transverse 

2-FillB2Gjc% 

2  Pis.  %  Longitudinal 

2-FillB  % 

4  Ls  6  x  0  x  %1 


Bott.  of  Anchors  Rods  3*^  rt  per  column) 
Scale  1  =  30' 


ASSUMED  LIVE  LOAD 

Floor.  Sj^Btem  Railway  Stringers,  etc.  -50  ton  electric  locomotive 
+2000  Ib.  p.l.f. 
"Impact  50^6 
Roadway  Stringers    •   6  ton  auto  truck  or  60  Ibs.per  sq.ft. 

Impact  25$ 

Sidewalk  Stringers  -    CO  Ibs.  per  sq.  ft.  No  Impact 
Trusses,  etc.    Railway  Loading  2000  Ibs.  uniform  load 

Simultaneous  roadway  and  sidewalk  loading  700  lbf.p.U. 
Cables,  etc.      Entire  bridge  loaded  with  50  Ibs.  per  sq.  ft. 
Wind  Load      For  Suspension  Bridge  25  Ibs.  per  sq.  ft.  Viaducgt  3^1bs. 
Temperature  Variation  of  ±  30°T. 


ASSUMED 
DEAD  LOAD 
Cables  4S 

Wrapping  S 

Bands,  etc.  4 

Suspenders  2 

Floor  System     740 
Trusses  1147 

Lateral  Bracing  315 
Flooring  000 

Railings 

Water  Main      J325 
TOTAL  p.l.f.    4050 


Sicle  Span 

: 

Chords 

Diagona  s 

Chords 

1-:: 

•;;.; 

4Lfi4xlx% 

2  WetB  i4xj^ 

u-1 

4  Ls  4x4x»/8 

2  V«b«  24x>6 

0-1 

4  Ls  4x4x% 

2  \Vflid-24x 

1-2 

4  Ls  Bx3Mx% 

1-:; 

5-7 
7.9 

4La4x4iM 

•• 

2-:i 

4LsBx.^x^ 

3-3 

4Ls4x^H 

:;-i 

4LsBx3^x^ 

,%7 

4  Ls  4xlx% 

y-n 

4-5 

4  Ls  6x:i^x% 

7-9 

.-. 

l-i: 

4  Ls  4x4*K 

£g 

9-11 

2  Web«  24< 

l.i.ir 

4L«4x4x% 

li-7 

]-!:: 

... 

i:,-r 

7-8 
0-9 

is.is 

Vi 

15-17 

9-  10 

17-10 

n-2 

1  L«  4x4x% 

2  Webu  24x^ 

i~.\ 

K.ii 

•>-l 

].}•- 

!i.V:: 

I  -ti 

•:-\: 

3-29 

•::.'< 

4  Lfi  4x4x^ 

;-i 

., 

!j-z; 

U10 

4Ls  4x4x% 

1-1: 

nj^H3? 

2WrtSTx 

-l: 
r^-n 

J^S 

r-Il. 

i;-r 

0.2 

2~:r 

4  Ls  4x4i?'i 

* 

6-lf 

7-» 

j.i; 

C-8 

4  Ls  4x4xJi 

«. 

a 

•• 

3.10 

ft. 

i  '-;•.' 

2Weta2i, 

-:-M 

]-!' 

8.18 

•<-•.'' 

•* 

!-'/(. 

8-27 

•  < 

FIG.  43. — Suspension  Bridge  at 
Span,  1113  feet  9  inches.     Design  by  I 


lowers 


(Expansion  Point  All  eub  struts  4  La  3^  x  8  x  %    D.L.  2^ 


•K 


2  Ls  6  i  3K  x  K  [-  2  Ls  C  x  3H  *  KijT 


%  £ 


2  Ls  3^  z  3  I  % 


All  truss  verticals  4  is  3}f  x  3  x  %      S.L  2H  : 


5        0        7        8        9       10      11       12      13      14      15      10      17       18      19      20      21      22      23       24      25      26    27 
64  Panel»-@  20f7H"=  1113  Vc.  c.  Tower* 


I  Expansion  Poit 


\ 


\ 


\ 


\ 


b  plates 

ASSTJMED  UNIT  STRESSES 
Floor  System:-      Tension  17,000  Ibs.  per  sq.  in. 

Compression  17,000-80  £ 
Towers,  etc:-          Tension  18.600  Ibs.  per  sq.  In. 

Compression  18,500-85-^ 
StiiTenin?  Truss:-  Tension  20.COO  Ibs.  per  sq.  in. 

Compression  20,000-9o£ 

For  combination  of  D.L.  L.L.,  1  and  W..L  increase 
unit  stresses  25~f> 

Suspenders:-          Tension  45.0IX)  Ibs.  per  sq.  in. 
Cables:-  Tension  05,0(10  Ibs.  per  sq.  In. 

Kote:- 

See  Sheet  2  of  2  for  Stress  Sheet  of  Approach  Span 

FLORIANOPOLIS  SUSPENSION  BRIDGE 

SANTA  CATHARINA,  BRAZIL 

Section  Sheet  of  Suspension  Span  1856*3  Long 

BYINGTON  &  SUNOSTROM.SAO  PAULO,  BRAZIL,  CONTRACTORS 
.  ROBINSON  &  STEINMAN,  U.S.A.  CONSULTING  ENGINEERS 
1  7)4  Sheet  1  of  2  Aug.  1,  1920 


at  Florianopolis,  Brazil.      (Type  25) 
H.  D.  Robinson  and  D.  B.  Steinman,  1920. 


(To  face  page  134.) 


TWO-HINGED   BRIDGE  WITH  SUSPENDED   SIDE  SPANS     135 

For  Dead  Load,  by  Eq.  (n),  the  horizontal  component  of 
cable  stress  is, 

H  =  ~=^-wl  =  $220  kips,     (i  kip  =  iooo  Ib.) 

of       o 

For  Live  Load,  by  Eq.  (125),  the  denominator  of  the  #-equa- 
tion  is, 


A]     I  A\j      I 

=  1.626+.  093+.  071  =  1.790. 

By  Eq.  (135),  the  horizontal  tension  produced  by  live  load, 
covering  all  three  spans,  will  be, 

T  *) 

H  =  -jgr(i  +  2ir3v)p/l  =  -  (i  .  0164)  (9300)  =  1050  kips. 

The  total  length  of  cable  between  anchorages  is  given  by 

Eq.  (154): 

L    f    .  8  2\  .    hf          .  8    m2  \ 

-=    i  +-n2  }  +2-1  sec  ai-\  ---  -  —  )  =  i  .027+  .  767  =  i  .  794. 

I      \      3    /       /  \  3  sec3  ail 

Then,  for  temperature,  by  Eq.  (156), 


**i 


/•97\T/ 

f2Nl 
Adding  the  values  found  for  H: 


_  3  (i  1  7  20)  (26200)  (i  .  794)  _ 

7      0\o  v  x  --  ~roO  Kips. 

(io8)2Xi.790 


D.  L.         3220  kips 
L.  L.          1050 
Temp.  80 


we  obtain,  Total  #=4350  kips  per  cable. 

The  maximum  tension  in  the  cable  is,  by  Eq.  (5), 

Ti  =  H  -  sec  0i  =  E(i .  08)  =  4700  kips. 
At  60,000  Ib.  per  sq.  in.,  the  cable  section  required  is: 
4700 -T- 60  =  78  sq.  in.  per  cable. 


136 


TYPICAL  DESIGN  COMPUTATIONS 


3.  Moments  in  Stiffening  Truss  —  Main  Span.  — 
Given: 

Live  Load  =p  —  1600  Ib.  per  lin.  ft. 

(All  values  given  and  calculated  are  per  truss.) 
With  the  three  spans  completely  loaded,  the  bending  moment 
at  any  section  x  of  the  main  span  is  given  by  Eq.  (140)  : 


Total  M  - 


Hence  only  9.1  per  cent  of  the  full  live  load  is  carried  by  the 
stiffening  truss.    Accordingly,  at  the  center, 


Total  M  —  .091  ~  =  +21.200  ft.  kips. 

o 


At  other  points,  the  values  of  M  are  proportional  to  the  ordinates 
of  a  parabola.     They  are  obtained  as  follows: 


Section 

Parabolic  Coefficient 

Total  M 

X 

7=° 

4X  o  X     i   =       o 

o 

O.I 

4X.i   X    .9  =    .36 

+  7,6oo 

O.2 

4X.2  X    .8  =    .64 

+  13,600 

0-3 

4X.3  X    -7   =    .84 

+17,800 

0.4 

4X.4  X    .6  =    .96 

+20,400 

0.45 

4X.45X    .55=    -99 

+  21,000 

o-S 

4X.5  X    .5   =1.00 

+21,  200  ft.  kips 

For  maximum  and  minimum  moments,  the  critical  points 
are  found  by  solving  Eq.  (142)  : 


•»•-=<>.  179-, 

y          y 


with  the  aid  of  Table  I  or  Fig.  12  or  13: 


TWO-HINGED  BRIDGE  WITH  SUSPENDED  SIDE  SPANS     137 


X 

y_ 

X 

C(k) 

i 

D(k) 

1 

I 

y 

0 

0 

(2.50) 

(-448) 

(.364) 

(-508) 

.1 

.036 

2.78 

.498 

.402 

.411 

.  2 

.064 

3.12 

•559 

.448 

.310 

•3 

.084 

3-57 

.640 

•  512 

.202 

•4 

.096 

4-17 

•747 

.603 

•095 

•45 

.099 

4-55 

•815 

.667 

.050+.  ooo 

•5 

.100 

5-op 

•895 

•755 

.oi6+.oi6 

•55 

•099  . 

5-55 

•995 

•950 

.  000+  .  050 

The  values  of  D(k),  found  from  Table  I  or  Fig.  13,  are  recorded 

N 
in  the  above  tabulation.     For  all  sections  from  #  =  —  •/  =  .  44  7/ 

4 
to  #  =  .553/,  there  are  double  values  of  D(k),  as  explained  under 

Eq.  (143')- 

The  values  of  the  minimum  moments  are  then  given  by 
Eq.  (143): 


Min.  M=  - 


and  the  maximum  moments  are  then  given  by  Eq.  (144) : 
Max.  M  =  Total  AT -Min.  M. 


Section 

T(-T) 

[            ] 

Minimum  M 

Total  M 

Maximum  M 

x 

7=0 

o 

(-541) 

o 

o 

0 

.1 

.09 

•444 

-16,700 

+  7,6oo 

+24,300 

.2 

.16 

•343 

—  22,900 

+13,600 

+36,500 

•3 

.  21 

•235 

—  20,600 

+17,800 

+38,400 

•  4 

•24 

.128 

—  12,800 

+  20,400 

+33,200 

•45 

.248 

-083 

-  8,600 

+21,000 

+29,600 

•5 

•25 

•065 

-  6,800 

+  21,200 

+28,  ooo  ft.  kips 

138  TYPICAL  DESIGN  COMPUTATIONS 

Dividing  the  maximum  and  minimum  moments  by  the  truss 
depth  (d  =  22.5),  we  obtain  the  respective  chord  stresses.  Adding 
the  temperature  and  wind  stresses  found  as  in  Example  i  and 
dividing  by  the  specified  unit  stresses,  we  obtain  the  required 
chord  sections. 

4.  Bending  Moments  in  Side  Spans.  — 

(p  =  i6oo  Ib.  per  lin.  ft.  per  truss). 

With  all  three  spans  completely  loaded,  the  bending  moment 
at  any  section  x\  of  either  side  span  is  given  by  Eq.  (141): 


Total  lfi=i/wi(/i-* 
Accordingly,  at  the  center, 


Total  Mi  =  .091  ^-  =  +2300  ft.  kips. 


There  are  no  critical  points  for  moments  in  the  side  spans.     The 
minimum  moments  are  given  by  Eq.  (145) : 

Min.  Mi  =  -yi-1    *  v-pl  =  —yi(i .  i24)pl. 

Accordingly,  at  the  center, 

Min.  MI=  —i 2(1. 1 24) (1730)  =  —23,400  ft.  kips. 
The  maximum  moments  are  given  by  Eq.  (146) : 

Max.  Mi  =  Total  Jlfi-Min.  MI. 
Accordingly,  at  the  center, 

Max.  Jl/i  =  +2300+23,400  =  +25,700  ft.  kips. 

At  other  sections,  the  moments  are  proportional  to  the  ordinates 
of  a  parabola: 


TWO-HINGED  BRIDGE  WITH  SUSPENDED   SIDE  SPANS     139 


Section 

Parabolic 
Coefficient 

Total  M  i 

Minimum  M  \ 

Maximum  M  j. 

—  =o 

0 

o 

o 

o 

k 

.1 

•36 

+  800 

—  8,400 

+  9,200 

.2 

.64 

+  1500 

—  15,000 

+  16,500 

•3 

.84 

+  1000 

—  19,600 

+  21,500 

•4 

.96 

+  22OO 

—  22,400 

+24,600 

•5 

i 

+  2300 

-23,400 

+25,  700  ft.  kips 

6.  Shears  in  Stiffening  Truss  —  Main  Span.  — 
(p  =  i6oo  Ib.  per  tin.  ft.) 

With  the  three  spans  completely  loaded,  the  shear  at  any 
section  x  of  the  main  span  is  given  by  Eq.  (147)  : 


Total  F-/-2*i- 


-  2*)[.  091]. 


The  shears  will  be  the  same  as  would  be  produced  by  loading  the 
span  with  9.1  per  cent  of  the  actual  load,  or  with  .091  pl= 157  kips. 

Total  F  =  i< 


I       X 

Section 

Total  V 

X 

r° 

o-5 

+79  kips 

.1 

0.4 

+63 

.2 

°-3 

+47 

•  3 

O.2 

+31 

•  4 

O.I 

+  16 

•5 

O 

o 

The  maximum  shears  are  given  by  Eq.  (149) 


140 


TYPICAL  DESIGN  COMPUTATIONS 


where  the  values  of  G(-\  are   taken  from  Table  I  or  Fig.  12, 
The  shears  are  obtained  as  follows :  (%pl  =  864  kips) . 


Section 

-(---} 

N\2       l) 

Kr) 

[] 

B)' 

Maximum  V 

X 

r° 

2.23 

.400 

.107 

i 

+  92  +  194 

.1 

1.79 

.482 

.136 

.81 

+  95+  96 

.2 

i  34 

.565 

•243 

.64 

+  134+    22 

•3 

0.89 

.647 

.424 

.49 

+  179 

•  4 

•45 

.726 

-673 

.36 

+  2IO 

-5 

0 

.800 

I  .OOO 

•25 

+216  kips 

maximum 


For  all  sections,  x<-ii  --  j  =  .277^  the  loading  for 
2  V       4  / 

shear  extends  from  the  given  section  x  to  a  critical  point  kl 
defined  by  Eq.  (150): 


. 

4     /—  2X  2X 

I 


The  values  C(k)  are  solved  for  k  and  G(k)  with  the  aid  of  Fig.  13: 


2X 

Section 

X-T 

C(k) 

1 

o 

s1* 

.446 

.362 

.1 

.8 

.558 

.448 

.2 

.6 

•744 

.600 

For  these  sections,  a  correction  is  to  be  added  to  the  values  of 
Max.   V  found  above.     This  additional  shear  is  given  by  Eq. 


Add.  F= 


TWO-HINGED  BRIDGE  WITH  SUSPENDED  SIDE  SPANS     141 


Section 

k 

(l-*)2 

8/1     x\ 

N\2      ij 

G(k) 

[] 

Add.  V 

X 

r° 

.362 

.407 

2.23 

.696 

-552 

+194  kips 

.1 

.448 

.305 

1.79 

.762 

•365 

+  96 

.2 

.600 

.160 

i-34 

.860 

.160 

+    22 

The  minimum  shears  are  then  given  by  Eq.  (153): 
Min.  V  =  Total  F-Max.  V. 


Section 

Total  V 

Maximum  V 

Minimum  V 

X 

r° 

+  79 

+286  kips 

—  207  kips 

.1 

+63 

+191 

-128 

.2 

+47 

+156 

-109 

•3 

+3i 

+179 

—  148 

•4 

+  16 

+  2IO 

-194 

•5 

0 

+216 

-216 

6.  Shears  in  Side  Spans.  — 

(p  =  1600  Ib.  per  lin.  ft,  /i  =  360  ft.) 

With  the  three  spans  completely  loaded,  the  shear  at  any 
sect-ion  xi  in  the  side  spans  will  be,  by  Eq.  (148), 


Total  Ki  -!#(/,-  a*.)  [i-  A  £ 


Since  h  =  i/,  these  shears  will  be  \  of  the  corresponding  values  in 
the  main  span. 


Section 

Total  Vi 

r° 

+26  kips 

.1 

+  21 

.2 

+  16 

•3 

+  10 

•  4 

+  5 

•5 

o 

142 


TYPICAL  DESIGN  COMPUTATIONS 


There  are  no  critical  points  for  shear  in  the  side  spans.     The 
maximum  shear  at  any  section  Xi  is  given  by  Eq.  (152) : 


Section 

—  («n>2)(  1 

N              \2       l\/ 

$ 

[] 

H)' 

Maximum  V\ 

x\ 

.0183 

.400 

•993 

i 

+286  kips 

.1 

.0147 

.482 

•993 

.81 

+  232 

.2 

.OIIO 

.565 

•994 

.64 

+183 

•3 

.0073 

.647 

•995 

•49 

+  140 

•4 

.0037 

.726 

•997 

•36 

+  103 

•5 

o 

.800 

i  .000 

•25 

+  72 

The  minimum  shears  in  the  side  spans  are  given  by  Eq.  (153') : 
Min.  Fi  =  Total  Fi-Max.  FI. 


Section 

Minimum  V\ 

Xi 

ir 

—  260  kips 

.1 

—  211 

.2 

-167 

•3 
•  4 

-130 
-  98 

•5 

-  72 

7.  Temperature  Stresses.  — 


The  stresses  in  the  main  span  from  temperature  variation 
are  figured  exactly  as  in  Example  i  (Type  2F),  using  Eqs.  (157) 
and  (158): 

Mt=-Ht-y, 

Vt=—  F,(tan  </>  —  tan  a)  .      (Here,  tan  a  =  o)  . 


TWO-HINGED  BRIDGE  WITH  SUSPENDED  SIDE  SPANS     143 

The  temperature  moments  in  the  side  spans  are  given  by  the 
formula : 

Mt=-Ht-yi, 

and  will  therefore  be  v(  =  i)  times  the  corresponding  main-span 
values. 


Section 

Parabolic 
Coefficient 

Mt 

—  =o 

o 

0 

k 

.1 

.36 

±350 

.2 

.64 

±610 

•3 

-84 

±810 

•4 

.96 

±920 

•5 

i 

±960  ft.  kips 

The  temperature  shears  in  the  side  spans  are  given  by  the 
formula  : 


F,  =  —  Fi 


—  tan  «i).     (Here,  tan  a\  =  .  267.) 


They  will  be  —  (  =  |)  times  the  corresponding  main-span  values. 
n 


Section 

tan  0i  —  tan  a\ 

vt 

Xi 

r° 

4»i=-i33 

±11  kips 

.1 

.106 

±  8 

.2 

.080 

=fc  6 

•3 

•053 

±  4 

-4 

.027 

=fc    2 

•5 

o 

O 

8.  Wind  Stresses. — The  wind  stresses  in  the  bottom  chords 
and  lateral  bracing  are  calculated  exactly  as  in  Example  i  (Type 

2F). 


144  TYPICAL  DESIGN  COMPUTATIONS 

The  assumed  wind  load  (  =  />  =  4oo  Ib.  per  lin.  ft.)  is  reduced 
by  the  fractional  amount,  at  span  center, 

wl* 

r_     -°I3^7          .I73 
p  wl*     I+.I73     °'147' 


(20  =  4770  Ib.  per  lin.  ft;  2  =  130  ft;  /=  1  24,000  in.2  ft2) 

Since  the  equivalent  uniform  value  of  r  is  f  of  the  mid-span  value, 
the  resultant  horizontal  load  on  the  span  is, 

^-|r=4oo-f(59)=35i  Ib.  per  lin.  ft 

Treating  this  value  as  a  uniform  load,  the  bending  moment  at 
the  center  is, 

Mw  =  ^-  =  ±51,000  ft.  kips. 

o 

Dividing  by  the  truss  width,  42.5  ft.,  we  obtain  the  chord 
stress  =±1200  kips  at  mid-span.     The  wind  stresses  at  other 
sections  will  be  proportional  to  parabolic  ordinates. 
The  end  shears  in  the  lateral  system  will  be: 


In  the  side  spans,  unless  they  exceed  1000  feet  in  length,  the 
reduction  in  effective  wind  pressure  may  be  neglected.  (In 

this  example,  -  would  amount  to  only  i  per  cent.)     Hence  the 

moments  and  shears  are  calculated  for  the  full  specified  wind 
load  of  400  Ib.  per  lin.  ft.,  acting  on  simple  spans  360  ft.  in  length. 

EXAMPLE  3 

Calculations  for  Towers  of  Two-hinged  Suspension 
Bridge  (Type  2S) 

1.  Dimensions. — The  bridge  is  the  same  as  in  Example  2. 

Each  tower  consists  of  two  columns  of  box  section,  stiffened 
with  internal  diaphragms,  and  rigidly  tied  together  with  trans- 
verse bracing  in  a  vertical  plane.  Each  tower  column  is  225  feet 


CALCULATIONS  FOR  TOWERS  145 

high  and  is  made  of  a  double  box  section,  42.5  inches  wide.  The 
other  dimension  (d),  parallel  to  the  stiffening  truss,  is  4  feet  at 
the  top,  increasing  to  9  feet  at  the  base.  The  walls  are  ij  inches 
thick  (made  up  of  f-inch  plates  and  corner  angles)  and  the 
vertical  transverse  diaphragm  is  f-inch  thick.  Splices  are 
provided  at  such  intervals  as  to  keep  the  individual  sections 
within  specified  limitations  of  length  or  weight  for  shipment. 
Horizontal  diaphragms  are  provided  at  splices  and,  in  general, 
at  lo-foot  intervals. 

The  tower  columns  are  battered  so  as  to  clear  the  trusses. 
They  are  42.5  feet  center  to  center  at  the  top  and  53.5  feet 
center  to  center  at  the  base. 

2.  Movement  of  Top  of  Tower.  —  The  towers  are  assumed 
fixed  at  the  base  and  the  cable  saddles  immovable  with  respect 
to  the  tower. 

The  maximum  fiber  stress  in  the  tower  columns  will  occur 
when  the  live  load  covers  the  main  span  and  the  farther  side 
span  at  maximum  temperature.  Under  this  condition  of  load- 
ing, the  top  of  the  tower  will  be  deflected  toward  the  main 
span,  as  a  result  of  the  following  deformations  : 

1.  The  upward  deflection  (A/i)  at  the  center  of  the  unloaded 
side  span. 

2.  The  elongation  of  the  cable  between  the  anchorage  and 
the  tower,  due  to.  the  elastic  strain  produced  by  the  applied 
loads. 

3.  The  elongation  of  the  cable  due  to  thermal  expansion. 
These  deformations  are  computed  as  follows: 

(Live  Load  =  pf  =  860  Ib.  per  lin.  ft.     H=  1040  kips.) 

i.  The  upward  deflection  A/i  is  found  by  considering  the 
unloaded  side  span  as  a  simple  beam  subjected  to  an  upward 
loading  equal  to  the  live-load  suspender  tensions  (Eq.  78)  : 


=  rjo  Ib.  per  lin.  ft.  per  truss, 


146  TYPICAL  DESIGN  COMPUTATIONS 

2.  The  elastic  elongation  of  the  cable  in  the  side  span  is,  by 
Eq.  (55), 

.  178(1  .077)  =  .  192  ft. 


3.  The  temperature  expansion  of  the  cable  in  the  side  span 
is,  by  Eqs.  (53)  and  (26), 


We  also  have: 


—  I  —  j  =  .156(1.  03  7)  =0.162  ft. 


i  .  8    m2 

-  =  sec«H  ---  -  —  =  1.037, 


A/i  3  sec3  «i 

AZi     16     n\ 

_L  =  --  -L_=o.i6o. 

A/i       3  sec3«i 

The  deflection  of  the  top  of  the  tower  is  then  given  by, 


Substituting  the  values  just  calculated,  we  obtain  the  maximum 
tower  deflection: 

(.428)  +  —?—  (.192+.  162)  =0.408  ft. 


1.037 

3.  Forces  Acting  on  Tower.  —  Considering  this  deflection  as 
produced  by  an  unbalanced  horizontal  force  P  applied  at  the 
top  of  the  tower,  this  force  may  be  calculated,  if  the  sectional 
dimensions  of  the  tower  are  known,  by  the  formula, 


X2 

In  the  present  case,  we  find  S  —•  Ax  =  1 740.     Hence, 
•p 

P=y0 =  17,200^0  =  7000  lb.  per  column. 

1740 

The  other  loads  acting  on  the  tower  are  the  vertical  reaction 
(V)  at  the  saddles,  and  the  end-shears  (Vi)  at  the  point  of  sup- 
port of  the  stiffening  truss.  The  saddle  reaction  is  given  by  the 
formula: 

V  =  2H  -  tan  0  =  2  X  4340  X  o .  4  =  +  3470  kips  per  column. 


CALCULATIONS  FOR  TOWERS 


147 


The  truss  reaction,  with  all  spans  loaded  and  maximum  tem- 
perature rise,  is, 

7i  =  (42+32)  +  (i4-f  n)  =  +99  kips  per  column. 
With  one  side  span  unloaded,  as  assumed  above, 

Vi  =  (45+32)  +  (n  -  140)  =  -  52  kips  per  column. 

The  inaccuracy  introduced  by  neglecting  this  uplift,  FI,  will  be 
on  the  side  of  safety;  therefore  the  column  need  be  figured  only 
for  the  horizontal  load  P  and  the  vertical  load  V. 

At  any  section  x  of  the  tower,  the  horizontal  deflection  (y) 
from  the  initial  vertical  position  of  the  axis  is  given  with  suffi- 
cient accuracy  by  the  equation  for  the  elastic  curve  of  the 
cantilever: 


4.  Calculation   of   Stresses.  —  The   resulting    extreme   fiber 
stresses  at  any  section  of  the  tower  will  be: 

r-      K-     A  ^  V  .Pxc.V(yQ-y)c 

Combined  Stress  =  —  H  —  —  H  —  J  _  Jt  . 
A       I  I 

The  computations   may  be  arranged  as  follows,  the  stresses 
being  figured  for  convenience  at  25-foot  intervals: 


Joint 

X 

yo-y 

A 

/ 

X2 

V 
A 

Pxc 

V(yo-y)c 

Combined 
Stresses 

I 

I 

I 

No. 

ft. 

ft. 

ft. 

sq.  in. 

in.2  ft.2 

Ib./sq.  in. 

0 

0 

0 

4.0 

280 

560 

0 

12,400 

0 

o 

12,400 

i 

25 

.068 

4.5 

295 

73° 

.86 

1  1,  800 

500 

700 

13,000 

2 

50 

•  134 

5-o 

310 

940 

2.66 

11,200 

900 

IIOO 

13,200 

3 

75 

.197 

5-5 

325 

1170 

6.40 

10,700 

I2OO 

1600 

i3,5oo 

4 

100 

.254 

6.0 

340 

1440 

6.94 

10,200 

1500 

1800 

13,500 

5 

125 

.305 

6-5 

355 

1740 

8.98 

9,800 

I6OO 

2000 

13,400 

6 

150 

.348 

7.0 

37° 

2080 

10.80 

9,400 

I800 

2OOO 

13,200 

7 

175 

.380 

7-5 

385 

2460 

12.42 

9,000 

1900 

2OOO 

12,900 

8 

200 

.400 

8.0 

400 

2880 

13-88 

8,700 

1900 

1900 

12,500 

9 

225 

.408 

9.0 

43° 

3850 

13.20 

8,100 

1800 

I7OO 

1  1,  600 

2  =  69.54 

148 


TYPICAL  DESIGN  COMPUTATIONS 


5.  Wind  Stresses. — To  the  above  tower  stresses  produced 
by  live  load  and  temperature,  must  be  added  the  stresses  due  to 
wind  loads. 

The  truss  wind  load  of  400  Ib.  per  lin.  ft.  produces  a  hori- 
zontal reaction  at  each  tower  of, 

360  -+400  —  =  266  kips. 

This  acts  at  Joint  No.  4,  (x  =  100). 

The  deflection  of  the  stiffening  truss  under  wind  load  pro- 
duces a  horizontal  reaction  at  the  top  of  each  tower  of  40  -;  and 
the  wind  on  the  surface  of  the  cables  produces  an  addition  to 

this  reaction  amounting  to  io( — (--);  hence  the  total  reaction 

\4     2/ 

at  the  tower  top  =  26  kips. 

The  wind  acting  directly  on  the  tower  is  assumed  at  25  Ib. 
per  sq.  ft.  of  vertical  elevation.  This  produces,  at  each  joint,  an 
equivalent  concentrated  load  of  25  X  (25^). 


Joint 

X 

d 

Wind 
Load 

Shear 

Moment 

Col- 
umn 
Dist. 

A 

Stress 
from 
W.L. 

Stress 
from 
L.L.+ 

Total 
Stress 

Temp. 

No. 

ft. 

ft. 

kips 

kips 

ft.  kips 

ft. 

sq.  in. 

Ib./sq.  in. 

Ib./sq.  in. 

Ib./sq.  in. 

0 

o 

4 

26+1 

o  " 

0 

42.5 

280 

o 

12.400 

12,400 

i 

25 

4-5 

3 

—  27 

-      675 

43-5 

295 

IOO 

13,000 

13,100 

2 

50 

5 

3 

-  30 

-  1,425 

44-5 

310 

IOO 

13,200 

13,300 

3 

75 

5-5 

3 

-  33 

-  2,250 

46.5 

325 

IOO 

13,500 

13,600 

4 

IOO 

6 

266+4 

-  36 

-  3,i5o 

48-5 

34° 

200 

i3,5oo 

13,700 

5 

125 

6-5 

4 

-306 

—  10,800 

49-5 

355 

600 

13,400 

14,000 

6 

150 

7 

4 

-310 

-18,550 

50-5 

370 

IOOO 

13,200 

14,200 

7 

175 

7-5 

5 

-314 

—  26,400 

5i-5 

385 

1300 

12,900 

14,200 

8 

2OO 

8 

5 

-319 

-34,375 

52.5 

400 

1600 

12,500 

14,100 

9 

225 

9 

3 

-324 

-42,475 

53-5 

430 

1800 

1  1,  600 

13,400 

The  bending  moments,  divided  by  the  column  distance,  give  the 
column  stresses,  and  these  divided  by  the  areas  give  the  unit 
stresses  from  wind  load. 


CABLE  AND  WRAPPING  149 

The  transverse  bracing  is  proportioned  to  resist  the  shears 
tabulated  above. 

EXAMPLE  4 
Estimates  of  Cable  and  Wrapping 

1.  Calculation  of  Cable  Wire.  —  Given  a  suspension  bridge 
in  which  a  cable  section  of  68  sq.  in.  is  to  be  provided.  To  find 
the  material  required  for  cables  and  wrapping.  Other  data  as 
in  Example  2. 

The  total  length  of  each  cable  is  given  by  Eq.  (154): 


3 
=  1080(1  .027)4-720(1  .034+  .003)  =  1110+746  =  1856  ft. 

To  this  must  be  added  43  ft.  of  cable  at  each  end,  between  end  of 
truss  span  and  anchorage  eyebars  (scaled  from  drawing)  ;  hence, 

Total  L  =  1856+86  =  1942  ft.  per  cable. 

No.  6  galvanized  cable  wire  will  be  used  =  o.  192  in.  diameter  = 
.029  sq.  in.  area.  Each  cable  consists  of  7  strands  of  336  wires 
each  =  23  5  2  wires  at  0.29  sq.  in.  =  68  sq.  in. 

Weight  of  No.  6  galvanized  wire  =  o.i  Ib.  per  ft. 

Total  cable  wire  =  2X2352  wires  at  1942  ft.  =9,150,000  lin.  ft. 

Total  weight  of  cable  wire  =  9,  150,000  ft.  at  o.i  Ib.  =  915,000  Ib. 

2.  Calculation   of   Cable   Diameter.  —  The   diameter  of   the 
cable  is  figured  as  follows:   The  area  of  a  strand  will  be  10  per 
cent  greater  than  the  aggregate  section  of  the  wires  composing 
it.     In  this  case  the  area  of  each  strand  will  be, 

no  per  centX-6T8-=  io-  7  sq.  in. 

The  corresponding  diameter  is  3.7  in.  The  cable  diameter  will 
be  3  strand  diameters  =  1  1  .  i  inches.  (Adding  the  thickness  of 
wrapping,  the  finished  cable  will  be  1  1  .4  inches  in  diameter.) 

3.  Calculation  of  Wrapping  Wire.  —  The  wrapping  consists 
of  No.  9  galvanized  wrapping  wire  (soft,  annealed),  weighing 
.06  Ib.  per  ft.     Deducting  lengths  of  cable  bands,  etc.,  there  will 
be  3250  ft.  of  cable  to  be  wrapped.    Since  the  wrapping  wire  is 
0.15  in.  diameter,  it  will  make  80  turns  per  lin.  ft.     The  diameter 


150  TYPICAL  DESIGN  COMPUTATIONS 

of  the  cable  is  n.i  inches,  hence  the  length  of  each  turn  will  be 
2.9  ft. 

Length  of  wrapping  wire  =  80  turns  at  2.9  ft. 

=  232  ft.  per  lin.  ft.  of  cable. 

Weight  of  wrapping  wire  =  232  ft.  at  .06  Ib. 

=  13.44  Ib.  per  lin.  ft.  of  cable. 

Total  wrapping  wire        =3250  ft.  of  cable  at  13.44  Ib. 

=44,000  Ib. 

4.  Estimate  of  Rope  Strand  Cables.  —  Instead  of  building 
the  cable  of  individual  wires,  manufactured  rope  strands  may 
be  used.  In  the  case  at  hand,  with  a  factor  of  safety  of  3,  there 
would  be  required  sixty-one  if  -inch  strands  per  cable.  These 
galvanized  steel  ropes  weigh  4.34  Ib.  per  ft.;  hence  the  total 
weight  hi  the  cables  would  be, 

2  X  1942  ft.  X6i  strands  at  4.34  Ib  =  1,030,000  Ib. 

The  diameter  of  the  resulting  cable  would  be  7  X  if  -in.  =  11.4  in., 
plus  the  wrapping.  (If  rope  strands  are  used,  it  should  be 
remembered  that  their  modulus  of  elasticity  E  is  only  about 
20,000,000,  as  compared  with  30,000,000  for  parallel  wire  cables.) 

EXAMPLE  5 

Analysis  of  Suspension  Bridge  with  Continuous  Stiffening  Truss 

(Type  OS) 

(See  Chap.  I..     Pages  53  to  63.) 

1.  Dimensions.  —  The  following  dimensions  are  given: 
/  =  Main  Span  =  40  panels  at  1  7'  7!"  =  705  ft.     (/'=/). 


/=  Cable  sag  in  Main  Span  =  74.  285  ft.     [n=j=  —  )• 

V       ^     9-5/ 


BRIDGE  WITH  CONTINUOUS  TRUSS  151 

/i  =  Cable-sag  in  Side  Spans  =  4.65  ft. 

d  =  Depth  of  Truss  =   15/0  at  towers,  10/833  at  center, 

11/346  at  ends. 
Width,  center  to  center  of  trusses  or  cables  =  27  ft. 

7  =  Mean  Moment  of  Inertia  in  main  span  =  1642  in.2  ft.2 

(per  truss). 

1 1  =  Mean  Moment  of  Inertia  in  side  spans  =2278. 

(i  =  ~r  =0.72  j. 
Ii  I 

A  =  Cable  Section  in  main  span  =  7  strands  of  282  wires  at 
0.192  in.  diameter  =57. 2  sq.  in.  per  cable. 

A 1  =  Cable  Section  in  side  spans  =  A,  +  2  strands  of  76  wires 
=  61 .6  sq.  in.  per  cable. 

tan  a  =  Slope  of  Cable  Chord  in  main  span  =    .026. 
tan  a\  =  Slope  of  Cable  Chord  in  side  spans  =0.5. 

e  =  Coefficient  of  Continuity  =  —  — —  =  o .  602 


2.  Stresses  in  Cables.  —  (All  values  per  cable). 
For  dead  load  (2^  =  2850  Ib.  per  lin.  ft.  per  cable),  the  hori- 
zontal tension  is  given  by  Eq.  (n)  : 

H  =  —=  ^  wl  =  2380  kips  per  cable. 

OJ  O 

The  denominator  for  other  values  of  H  is  given  by  Eq.  (203)  : 


With  the  live  load  (p  =  *]$o  Ib.  per  lin.  ft.  per  cable)  covering  the 
main  span,  Eq.  (206)  gives, 

=  6io  kips  per  cable. 


N 


152  TYPICAL  DESIGN  COMPUTATIONS 

With  the  live  load  (pi  =  750  Ib.  per  lin.  ft.  per  cable)  covering 
both  side  spans,  Eq.  (207)  gives, 


The  total  length  of  cable  is  given  by  Eq.  (154)  : 

_\     I>8 


Substituting  this  value  in  Eq.  (214),  we  obtain  the  cable  tension 
produced  by  temperature  variation  (/  =  ±60°  F.,  E^t—  11,720)  : 


Ht~  ~  f2  N  I  ==T=47  kips  per  cable- 

Combining  the  values  for  dead  load,  main-span  live  load,  and 
fall  in  temperature,  we  obtain, 

Max.  H  =  303  7  kips  per  cable. 

In  the  main  span,  the  maximum  slope  of  the  cable  is  tan  0 
=  tan  a +4^  =  447;  sec  0  =  i. 096.  For  this  slope,  the  stress 
in  the  cable  is,  by  Eq.  (5), 

Max.  T  =  H  •  sec  0  =  3330  kips. 

At  60,000  Ib.  per  sq.  in.,  the  cable  section  required  is  3330-^60 
=  55.5  sq.  in.  (The  section  provided  is  A  =  57.2  sq.  in.) 

In  the  side  spans,  the  maximum  slope  of  the  cable  is  tan  0i 
=  tan  ai  +4ni  =  .605;  sec  <i  =  1.1 7.  For  this  slope,  the  stress 
in  the  cable  is,  by  Eq.  (5), 

Max.  Ti=H-  sec  01=3550  kips. 

At  60,000  Ib.  per  sq.  in.,  the  cable  section  required  is  3550-7-60 
=  59.2  sq.  in.  (The  section  provided  is  A  1  =  61.6  sq.  in.) 

3.  Influence  Line  for  H. — For  a  concentration  P  traversing 
the  main  span,  the  values  of  H  are  given  by  Eq.  (204) : 


Taking  the  values  of  B(k)  from  Table  I  or  Fig.  12,  we  obtain 
the  following  main-span  influence  ordinates  for  H: 


BRIDGE  WITH  CONTINUOUS  TRUSS 


153 


Load  Position 

TT 

Ordinate  — 

k=o 

0 

0.  I 
0.2 

°-3 

0.4 

•387 
.948 
1.482 
1.860 

o-5 

2.OOO 

For  a   concentration  PI    traversing   either   side   span,   the 
values  of  H  are  given  by  Eq.  (205)  : 


where  k\l\  is  measured  from  the  free  end  of  the  span.  Substi- 
tuting the  values  of  B(ki),  we  obtain  the  following  side-span 
influence  ordinates  for  H  : 


Load  Position 

II 

Ordinate  — 

Pl 

*i  =  o 

0 

0.2 

—  .048 

0.4 

-.085 

0.6 

—  .100 

0.8 

—  .078' 

I.O 

0 

4.  Bending  Moments  in  Main  Span. — The  bending  moments 
will  be  obtained  by  the  method  of  unit  loads  applied  at  succes- 
sive panel  points,  using  Eq.  (182): 


In  this  case,  ^=44.7  ft.,  and  the  values  of   (y—ef)   are  as 
follows : 


154 


TYPICAL  DESIGN  COMPUTATIONS 


Panel  Point 

* 

7 

y-ef 

No.  20 

0 

-44  .7  ft. 

16 

.1 

-17.9 

12 

.2 

+    2.8 

8 

•3 

+  17-7 

4 

•4 

+  26.4 

o 

•5 

+  29.6 

NOTE.  —  In  this  bridge,  the  panel  points  were  numbered  consecutively  in  both 
directions,  starting  with  No.  o  at  the  middle  of  the  span;  No.  20  is  at  the  towers, 
and  No.  30  at  the  free  ends. 

The  value  of  H  is  a  constant  for  each  load  position  and  is  taken 
from  the  influence  table  figured  above. 

For  each  load  position,  the  moments  M\  and  M  2  at  the  towers 
are  given  by  Eqs.  (190)  and  (191): 


The  bending  moment  M  '  at  the  section  carrying  the  load  is, 


Using  the  three  above  equations,  we  obtain  the  following  con 
trolling  values  of  M'  for  a  unit  load  P=i. 


Load  Position 

If  i~at  Near  Tower 

M'  at  Load 

Mt  at  Far  Tower 

£  =  o 

o 

o 

o 

.1 

-46.6 

+  20.5 

-  9-7 

.2 

-75-o 

+47-7 

-25-4 

•3 

-87.7 

+  73-9 

—44.2 

•4 

-87.7 

+Qi  3 

-62.8 

•5 

O 

—  75.4 

+97-8 

-78.4 

BRIDGE  WITH  CONTINUOUS  TRUSS 


155 


These  values  give  the  three  vertices  of  the  equilibrium  triangle; 
and,  for  each  load  position,  the  values  of  Mr  for  other  sections 
may  be  tabulated  by  straight-line  interpolation.  Subtracting 
from  each  value  of  M'  the  corresponding  value  of  H(y— ef),  we 
obtain  the  unit-load  bending  moments  M.  A  typical  tabula- 
tion for  this  computation  is  as  follows: 

UNIT  LOAD  AT  PANEL  POINT  12.     (£  =  0.2).     (#=.948) 


Panel  Point 

M' 

H(y-ef] 

M 

No.  20 

-75-0* 

-42.4 

-32-6 

16 

-i3-7 

-17.0 

+  3-3 

12 

+47-7* 

+  2.7 

4-45-Q 

8 

+38.6 

+  16.8 

+  21.8 

4 

+  29.4 

+  25.1 

+  4-3 

o 

+  20.3 

+  28.1 

-   7-8 

4 

+  11.  2 

+  25.1 

-13-9 

8 

+     2.0 

+  16.8 

—  14.8 

12 

-     7.2 

+  2.7 

-  99 

16 

-I6.3 

-17.0 

+  0.7 

20 

-25-4* 

-42.4 

+  17.0 

With  the  left  side-span  completely  loaded  (unit  load  at  each 
panel  point),  Eqs.  (197),  (198)  and  (207)  give: 

2ir3(i+ir) 
'     ' 


,,  i 

Mi  =  -^— 
4 


=  -  .ooi453/>i/2=  -41  .o 


The  resulting  bending  moments  in  the  main  span  will  be,  by 
Eq.  (182), 


and  are  obtained  by  a  tabulation  similar  to  the  one  above. 

The  influence  values  of  M  obtained  in  the  series  of  tabula- 
tions just  described  may  be  summarized  as  follows  (only  every 
fourth  panel  point  shown  here)  : 


156  TYPICAL  DESIGN  COMPUTATIONS 

INFLUENCE  VALUES  OF  M  FOR  UNIT  LOADS 


M  at  Panel  Point 

Load  Position 

20 

16 

12 

8 

4 

O 

Panel  point  No.  20  

0 

0 

O 

o 

0 

0 

16  

-   29.3 

+  27.4 

+  16.0 

+     6.8 

+       O.I 

-     4-6 

12  

-  32.6 

+  3.3 

+  45-o 

+    21.8 

+     4-3 

-     7-8 

8  

-   21.3 

7-4 

+  15-4 

+  47-Q 

+   17-  1 

4-i 

—     46 

—     06 

—     22 

+   i*  8 

+  41  8 

4-   10  6 

o  

+     II  -0 

7-4 

-13-5 

-     8.1 

+     9-4 

+  38.6 

4  

+  20.3 

3-9 

-   16.8 

-   18.8 

-     95 

+   10.6 

8  

+     22.2 

-     0.8 

-   14.8 

—     20.  I 

—   16.4 

4-i 

12  

+     l8.7 

+     0.7 

-     9.9 

-     14.8 

-   13-9 

7-8 

16  

+     7-6 

+     0.4 

4-3 

-     6.8 

-     6.7 

4.6 

20  

0 

0 

o 

o 

o 

o 

Left  side  span  

+  12.3 

+   23.7 

+  3i-3 

+    35  -O 

+    34-0 

+  31  -° 

Right  side  span  

-  46.3 

-23.1 

-     39 

+  ii.  6 

+    23.1 

+  31-0 

Maximum  M  

+323-1 

+  I35-4 

+3I7-4 

+379-8 

+323.6 

+  275-8 

Minimum  M  

-411.4 

-139.8 

-245-1 

—  270.6 

-184.3 

-127.4 

Max.  M  is  the  summation  of  all  positive  influence  values,  and 
Min.  M  is  the  summation  of  all  negative  influence  values. 

These  results  are  multiplied  by  the  panel  load  (P  =  —  =13.22 

\       40 

kips)  to  obtain  the  bending  moments  in  foot-kips ;  and  the  latter 
values  are  divided  by  the  truss  depths  at  the  respective  panel 
points  to  obtain  the  chord  stresses  in  kips. 

The  temperature  moments  are  given  by  Eq.  (215): 

Mt=-Ht(y-ef), 

where  Ht=^F4fj  kips;  the  values  of  (y  —  ef)  have  been  tabulated 
above.  These  moments  are  combined  with  the  live-load 
moments  as  the  specifications  may  prescribe. 

6.  Shears  in  Main  Span. — The  shears  in  the  main  span  are 
given  by  Eq.  (188): 


F=Fo+-- 


- — #(tan  0  —  tan  a)  =  V — #(tan  0  —  tan  a) . 


BRIDGE  WITH  CONTINUOUS  TRUSS 


157 


The  method  of  unit  loads  will  be  used.     The  values  of  H,  M 
and  Mi  have  been  calculated  above  for  different  load  positions. 


M-2-Mi 

V 

V 

Load  Position 

Fo 

(to  left  of  load) 

(to  right  of  load) 

H 

I 

Panel  point  No.  20  ... 

1  .0 

0 

+1.0 

0 

o 

16... 

•9 

.020 

+   -920 

-.080 

•387 

12.  .. 

.8 

.067 

+   .867 

-•133 

.948 

8... 

•  7 

.092 

+   -792 

-.208 

1.482 

4... 

.6 

.066 

+   .666 

-•334 

1.860 

o.  .  . 

•  5 

o 

+   -500 

—  .500 

2.OOO 

The  value  of  (tan  </>  —  tan  a)  decreases  uniformly  from  4n  =  .42 
at  the  left  tower  (Panel  Pt.  No.  20)  to  o  at  the  center  (Panel  Pt. 
No.  o)  and  to  —  ^n=  —.42  at  the  right  tower  (Panel  Pt.  No.  20). 
Substituting  these  values  in  the  above  formula  for  V,  an 
influence  table  for  shears  is  constructed,  similar  to  the  preceding 
influence  table  for  moments.  (Only  every  fourth  panel  point 
shown  here) : 

INFLUENCE  VALUES  OF  M  FOR  UNIT  LOADS 


V  at  Panel  Point 


L,oaa  rosmon 

20 

16 

12 

8 

4 

0 

Panel  point  No.  20  

+  1  .OO 

o 

O 

0 

o 

o 

16  

+     .76 

—     .21 

-     .18 

—    .  15 

—    .  ii 

-    .08 

12  

+    -47 

+    -55 

-    -37 

—    .29 

—     .  21 

—    -13 

8  

+   -17 

+    .29 

+    .42 

-    .46 

-    -33 

—     .21 

4  

—    .11 

+   -05 

+     .20 

+    -36 

-    -49 

-    -33 

o  

-    -34 

—    .17 

0 

+    .16 

+    -34 

±   -50 

4. 

—    .4C 

—    .  29 

—    •  J4 

+     .02 

+    -17 

+     33 

8  

-    -41 

-    .29 

-    .16 

.04 

+    -09 

+     -21 

12  

-    .27 

-    .19 

—    .11 

-     -03 

+    -05 

+  .13 

16  

-    .08 

-    -05 

—    .02 

+     -01 

+    -05 

+  .08 

20  

O 

o 

o 

0 

o 

o 

Left  side  span  

+    -35 

+    -30 

+    -24 

+  .19 

+   -13 

+  .08 

Right  side  span 

+     19 

+     14 

+    .08 

+  .03 

—     03 

—    08 

Maximum  V  

+8.02 

H~4-  74 

+3  -42 

+  2.94 

+3  -43 

+4.56 

Minimum  V  

-6.58 

-4.58 

-3-6o 

-3.38 

-4.12 

-4.56 

158 


TYPICAL  DESIGN  COMPUTATIONS 


Max.  V  is  the  summation  of  all  positive  influence  values,  and 
Min.  V  is  the  summation  of  all  negative  influence  values.     These 

results  are  multiplied  by  the  panel  load  [P  =  —  =13.22  kips) 

\        40  / 

to  obtain  the  vertical  shears  in  kips. 

The  temperature  shears  are  given  by  Eq.  (217): 

Vt=  -#<(tan  0-tana). 


The  shears  are  then  multiplied  by  the  respective  secants  of 
inclination,  to  obtain  the  stresses  in  the  web  members  of  the 
stiffening  truss. 

6.  Bending  Moments  in  Side  Spans.  —  The  bending  moments 
in  the  side  spans  are  obtained  by  the  method  of  unit  loads,  using 
Eq.  (183): 


h  V      /i 

For  loads  in  the  main  span,  MQ  =  O,  and  the  values  of  MI  and  71/2 
are  the  same  as  calculated  above.  For  the  far  side  span  com- 
pletely loaded,  MQ  =  O,  and  the  value  of  MI  is  the  same  as  the 
value  of  Mi  calculated  above.  For  unit  loads  (P—i)  in  tLe 
given  side  span,  the  moment  M\  is  given  by  Eq.  (194) : 

2ir2(i+ir)(ki-ki3) 


Mi=-Pl 


and 


The  values  of  H  will  be  the  same  as  calculated  above.  Accord 
ingly,  we  have  the  following  values  for  a  unit  load  (P=i)  tra 
versing  the  side  span. 


Load  Position 

kl 

if 

H 

Panel  point  No.  20  

I  .O 

O 

0 

22  

.8 

+24.4 

-.048 

24  

.6 

+38.5    1    -.085 

26  

•4 

+40.1 

—  .100 

28  

.  2 

+27.6 

-.078 

30  

0 

0 

0 

BRIDGE  WITH  CONTINUOUS  TRUSS 


159 


The  values  of  y'  —  y\  ~--ej  are  as  follows: 
/i 


Panel  Point: 


20 


x\ 

-=          1.0 

k 


22  24 

.8  .6 


26  28     30 

.4  .20 


/=-44.7     -35-8     -26.8     -17.9     -8.9      o 
Substituting  the  various  values  in  the  equation: 

M  =  M'-H-y', 

we  obtain  the  following  influence  table  for  side-span  moments 
(only  every  second  panel  point  shown  here) : 

INFLUENCE  VAIUES  OF  M  FOR  UNIT  LOADS 


H  at  I 

'and  Pcii: 

t 

Load  Position 

20 

21 

22 

24 

26 

28 

30 

Panel  point  No.  20  . 

0 

0 

0  „ 

0 

o 

O 

o 

22  . 

-     6.8 

f     8.0 

+     22.8 

+   17-2 

+  ii.  6 

+     5-8 

0 

24. 

—   10.  i 

+     1.6 

+     13-3 

+  36.6 

+  24.6 

+  12.3 

0 

26. 

—     10    0 

-     1.7 

+     6.4 

+   22.7 

+  38-8 

+  19-5 

o 

28. 

-   6.7 

-     2.3 

+     1.9 

+   10.5 

+  18.9 

+  27.1 

0 

3°- 

o 

0 

0 

0 

o 

o 

0 

Maximum  M  

0 

+  27 

+  £8 

+  179 

+  193 

+  130 

0 

Minimum  M  

—   70 

-     8 

o 

o 

o 

o 

o 

Far  side  span  

—    12 

-     9 

-     7 

4 

—       2 

—     I 

0 

Main  span  

+  311 

+  226 

+  167 

+   71 

+    14 

+     4 

o 

-365 

-356 

-349 

-318 

-247 

-153 

0 

Total  Maximum  M. 

+  311 

+  253 

+  255 

+  250 

+  207 

+  134 

0 

Total  Minimum  M. 

-447 

-373 

-356 

-322 

-249 

-154 

0 

The  above  results  are  to  be  multiplied  by  the  panel  load 

(p=—  =  — =13.22  kips )   to  obtain  the  maximum  and  mmi- 
40     10  *y 

mum  bending  moments  in  the  side  span. 


160  TYPICAL  DESIGN  COMPUTATIONS 

The  temperature  moments  are  calculated  by  Eq.  (216): 


where  Ht  —  T47  kips. 

7.  Shears   in   Side   Spans.  —  The  left  side-span   shears   are 
calculated  by  Eq.  (189)  : 


V= 


\  -tf  /tan  01- 
h  /        \  . 


-i  =  V'-K-H. 


ef 
At  the  tower  (Panel  Point  20),  K=  —  4^1—  j-  =  —  .105—  .254 

k 
=  —  .  359;  and  the  value  of  K  diminishes  uniformly  to,  K= 


—  —  =  +  .  105  ~~  •  254  =  ~  •  I49  at  the  free  end  (Panel  Point  30). 
k 

Substituting  in  the  above  formula  the  known  values  of  H, 
MI  and  K,  we  obtain  the  following  table  of  influence  ordinates 
for  side-span  shears  (only  every  second  panel  point  shown  here) : 

INFLUENCE  VALUES  OF  V  FOR  UNIT  LOADS 


V  at  Panel  Point 

Load  Position 

20 

22 

24 

26 

28 

30 

Panel  point  No.  20  

o 

0 

0 

o 

0 

o 

22  

-     .80 

-     .81 

+    -19 

+    .19 

+  .18 

+    .18 

24  

-     .60 

-     .61 

-    .61 

+    -38 

+  .38 

+    -38 

26 

—       4O 

—       41 

—     41 

—     41 

+       ^Q 

+     58 

28  

—     .20 

—     .21 

—     .21 

—     .21 

—     .21 

+    -79 

30  

O 

0 

0 

O 

O 

o 

Maximum  V  

o 

+  0.1 

+0.6 

+  i-4 

+  2.7 

+4-4 

Minimum  V 

—  4    "? 

—  V  7 

—  2.2 

—  i.i 

—  o  3 

o 

Far  side  span  

+0.3 

+0.3 

+0.3 

+0.2 

+0.2 

+0.2 

Main  span  

+5-4 

+3-8 

+  2.1 

+  1.2 

+0-3 

O 

-0.9 

—  i.i 

-1.6 

-2-5 

-3-4 

-5-i 

Total  Maximum  V  

+5-7 

+4-2 

+3-0 

+  2.8 

+3-2 

+4-6 

Total  Minimum  V  

-5-4 

-4-8 

-3-8 

-3-6 

-3-7 

-5-i 

BRIDGE  WITH  CONTINUOUS  TRUSS 


161 


These  results  are  to  be  multiplied  by  the  panel  load  (P=i$.22 
kips)  to  obtain  the  maximum  and  minimum  shears  in  kips. 


FIG.  44. — Design  of  Anchorage. 


For  the  right  side  span,  the  shears  will  be  the  same,  with  the 
signs  changed. 


162  TYPICAL  DESIGN  COMPUTATIONS 

The  temperature  shears  are  given  by  Eq.  (218): 

-        =  -K-Ht. 


The  shears  are  then  multiplied  by  the  respective  secants  of 
inclination,  to  obtain  the  stresses  in  the  web  members. 

EXAMPLE  6 
Design  of  Anchorage 

1.  Stability  against  Sliding.  —  The  outline  of  a  design  for  a 
reinforced  concrete  anchorage  is  shown  in  Fig.  44. 

The  principal  forces  acting  are  the  cable  pull,  and  the  weight 
of  the  anchorage  (including  any  superimposed  loads).  In  the 
case  at  hand,  the  cable  pull  =  H  -sec  0  =  7800  kips.  The  weight 
of  the  anchorage  and  the  superimposed  loads  is  30,000  kips. 
This  weight  is  represented  in  the  diagram  as  a  vertical  force 
drawn  through  the  center  of  gravity  of  the  anchorage  and  applied 
loads.  By  a  parallelogram  of  forces,  the  total  resultant  is  found, 
amounting  to  29,000  kips.  If  its  inclination  from  the  vertical  is 
less  than  the  angle  of  friction,  the  anchorage  is  safe  against 
failure  by  sliding. 

2.  Stability  against  Tilting.  —  The  resultant  is  prolonged  to 
intersection  with  the  plane  of  the  base,  and  its  vertical  com- 
ponent  (V  —  28,000  kips)   is   considered   as   an  eccentric  load 
applied  at  the  point  of  intersection.     The  toe  and  heel  pres- 
sures are  given  by, 

.     V     Vec 

P=A±-T> 

where  A  is  the  area  of  the  base  (sq.  ft.),  /  is  its  moment  of  inertia 
about  the  neutral  axis  (ft.4),  e  is  the  distance  (ft.)  of  the  resultant 
V  from  the  neutral  axis,  and  c  is  the  distance  (ft.)  from  the 
neutral  axis  to  the  respective  extreme  fiber.  We  thus  obtain, 
for  the  case  at  hand,  a  toe  pressure  of  10.6  kips  per  sq.  ft.  and  a 
heel  pressure  of  1.8  kips  per  sq.  ft.  The  allowable  foundation 
pressure  was  6  tons  per  sq.  ft.,  so  the  anchorage  figures  safe 
against  settlement  or  overturning. 


CHAPTER  IV 
ERECTION  OF  SUSPENSION  BRIDGES 

1.  Introduction. — The    erection    of    suspension    bridges    is 
comparatively  simple,  and  is  free  from  dangers  attending  other 
types  of  long  span  construction. 

The  normal  order  of  erection  is:  substructure,  towers  and 
anchorages,  footbridges,  cables,  suspenders,  stiffening  truss  and 
floor  system,  roadways,  cable  wrapping. 

The  cables  are  the  only  members  requiring  specialized  knowl- 
edge for  their  erection.  The  other  elements  of  the  bridge,  for 
the  most  part,  are  erected  in  accordance  with  the  usual  field 
methods  for  the  corresponding  elements  of  other  structures. 

2.  Erection  of  the  Towers. — The  erection  of  the  towers  may 
proceed  simultaneously  with  the  construction  of  the  anchorages. 

In  the  case  of  the  Manhattan  Bridge,  the  tower  (Fig.  45)  con- 
sists of  four  columns  supported  on  cast-steel  pedestals  resting 
on  base  plates  set  directly  on  the  masonry  pier.  The  sections 
of  the  pedestals  (weighing  up  to  40  tons)  were  delivered  by  light- 
ers and  lifted  by  their  derricks  to  the  pier- tops;  they  were  rolled 
into  position  on  cast-steel  balls  placed  on  the  bed  plate,  and  then 
jacked  up  to  release  the  balls. 

The  tower  columns  were  erected  by  the  use  of  ingenious 
derrick  platforms  (one  for  each  pair  of  columns)  adapted  to 
travel  vertically  up  the  tower  as  the  erection  proceeded  (Fig.  45). 
Each  platform  (21  feet  by  34  feet)  projected  out  from  the  face 
of  the  tower  on  the  shore  side  and  was  supported  by  two  bracket- 
struts  below.  The  tipping  moment  was  resisted  by  two  pairs  of 
rollers  or  wheels,  one -at  each  column,  engaging  vertical  edges 
of  the  projecting  middle  portion  of  the  column,  the  upper  wheel 
being  on  the  river  side  and  the  lower  wheel  on  the  shore  side. 
The  vertical  support  was  furnished  by  hooks  engaging  the  pro- 

163 


164 


ERECTION  OF  SUSPENSION  BRIDGES 


jecting  gusset  plates  of  the  bracing  system.  A  stiff-leg  derrick 
with  4 5 -foot  steel  boom  was  mounted  at  the  middle  of  the  inner 
side  of  the  platform,  being  braced  back  to  the  outer  corners  of 
the  platform.  With  this  derrick  the  sections  of  the  tower 
(weighing  up  to  62  tons)  were  lifted  from  the  top  of  the  pier  and 
set  in  place,  the  material  having  been  transferred  from  scows  to 
the  pier  by  floating  derricks.  When  a  full  section  had  been 
added  to  the  tower,  blocks  were  fastened  to  the  top  and  falls 


FIG.  45. — Manhattan  Bridge.     Erection  of  Towers. 

(See  Fig.  35,  page  97). 

attached  to  the  derrick  platform  by  which  it  then  lifted  itself  to 
the  next  level. 

For  purposes  of  handling  and  erection,  each  column  was 
divided  by  transverse  and  longitudinal  field  splice  joints  into 
sections  of  convenient  size.  The  transverse  joints  were  12  feet 
to  27!  feet  apart,  and  were  staggered  to  break  joint.  Where  the 
three  longitudinal  sections  changed  to  two,  shim  plates  were 
used  to  level  off.  The  riveting  of  the  field  splices  (with  i-inch 
rivets)  was  kept  several  sections  back  of  the  erection  work  in 


TOWER  ERECTION  165 

order  to  give  opportunity  for  the  transverse  joints  to  come  to 
full  bearing. 

Each  tower  column  was  finished  with  a  cap  section  (52  tons) 
upon  which  was  set  the  saddle  (15  tons). 

In  addition  to  the  two  traveling  derricks,  the  following 
equipment  was  required  for  the  erection  of  each  tower:  two 
hoisting  engines  on  the  pier;  one  stiff-leg  derrick  (lo-tons, 
6o-foot  boom)  on  the  pier  between  the  tower  legs,  used  in  the 
assembly  of  the  traveling  derricks;  two  large  storage  scows 
moored  to  the  pier,  supplying  the  respective  traveling  derricks; 
a  power  plant  on  shore  with  two  50-H.P.  horizontal  boilers,  a 
steam  turbine  blower  for  forced  draft,  and  an  air  compressor; 
30  pneumatic  riveting  hammers;  6  pneumatic  forges. 

The  force  at  each  tower  consisted  of  a  hundred  men,  including 
six  riveting  gangs.  Riveting  scaffolds  were  erected  around  the 
tower  for  field  riveting,  and  were  provided  with  stairs  and 
safety  railings.  The  erection  record  was  2000  tons  of  steel  at 
one  tower  in  sixteen  working  days. 

Figure  46  shows  the  completed  tower,  282  feet  high  above 
the  masonry,  and  weighing  12,500,000  pounds. 

For  the  Manhattan  tower  of  the  Williamsburg  Bridge, 
a  stationary  derrick  on  the  approach  falsework  was  used  to  erect 
the  steel  up  to  roadway  level;  the  erection  was  then  completed 
by  two  stiff-leg  derricks  mounted  on  a  timber  tower  built  up  on 
the  cross-girder  between  the  two  tower  legs.  (The  completed 
tower  is  shown  in  Fig.  57.) 

For  smaller  bridges,  the  towers  may  be  erected  by  gin-pole 
or  by  stationary  derrick  alongside.  For  the  suspension  bridge 
at  Kingston,  N.  Y.  (H.  D.  Robinson,  Chief  Engineer),  a  guyed 
derrick  with  95-foot  steel  boom  was  set  up  on  a  square  timber 
tower  80  feet  high,  for  the  erection  of  each  steel  tower;  the  same 
derricks  later  erected  the  adjoining  panels  of  the  stiffening  truss 
(Fig.  56). 

3.  Stringing  the  Footbridge  Cables. — The  first  step  in  cable 
erection  consists  in  establishing  a  connection  between  the  two 
banks.  Various  methods  have  been  used  since  prehistoric  times, 
when  the  first  thread  was  fastened  to  an  arrow  and  shot  across 


166 


ERECTION  OF  SUSPENSION  BRIDGES 


from  bank  to  bank.  In  building  the  Niagara  bridge,  a  kite  was 
used  to  take  the  first  string  across  the  gorge;  at  other  places,  a 
light  rope  is  drawn  across  with  a  rowboat. 

In  the  erection  of  the  Brooklyn  Bridge,  a  f-inch  wire  rope 
was  first  laid  across  the  East  River  by  means  of  a  tugboat  and 
scow,  and  then  raised  to  position.  With  another  line  taken 
over  in  the  same  manner,  an  endless  rope  was  made,  and  this 
was  used  for  hauling  over  the  remaining  traveler  ropes  and  an 


FIG.  46. — Manhattan  Bridge.    Erection  of  Footbridges. 

auxiliary  if -inch  carrier  rope;  the  latter  served  to  carry  the  load 
of  the  footbridge  cables  and  cradle  cables  (2§  and  2§  inches 
diameter)  when  these  were  hauled  across  the  river. 

For  the  Manhattan  Bridge,  sixteen  if -inch  wire  ropes  were 
swung  between  the  towers  in  four  groups  of  four  (Fig.  46) .  One 
group  (to  make  a  single  footbridge  cable)  was  taken  across  at  a 
time.  The  four  reels  were  mounted  on  a  scow  brought  along- 
side one  of  the  towers,  A.  The  end  of  each  rope  was  unreeled 
and  hauled  up  over  a  temporary  cast-iron  roller  saddle  mounted 


ERECTION  OF  FOOTBRIDGES  167 

on  top  of  the  tower,  and  thence  carried  back  to  the  anchorage  A, 
where  it  was  made  fast.  Then  the  scow  was  towed  across  the 
river,  laying  the  ropes  along  the  bottom,  to  the  opposite  tower  B. 
The  remainder  of  each  rope  was  then  unreeled  and  coiled  on  the 
deck  of  the  scow.  Then,  while  all  river  traffic  was  stopped  for  a 
few  minutes,  the  free  end  of  each  rope  was  hauled  up  by  a  line 
to  the  top  of  Tower  B,  over  a  roller  saddle  and  thence  to  the 
Anchorage  B;  the  middle  of  the  rope,  or  bight,  rose  out  of  the 
water  during  this  operation,  and  came  up  to  the  desired  position. 
After  being  made  fast  at  the  Anchorage  B,  the  ropes  were  socketed 
and  drawn  up  to  the  precise  deflection  desired,  as  determined  by 
levels.  Each  group  of  four  ropes  then  formed  a  temporary 
cable  for  the  support  of  the  footbridges  (Fig.  46). 

The  footbridge  ropes  for  the  Williamsburg  Bridge  were  strung 
in  the  same  manner  as  for  the  Manhattan  Bridge,  except  that 
three  were  laid,  instead  of  four,  at  each  trip  of  the  flatboat 
across  the  river. 

4.  Erection  of  Footbridges. — The  next  step  is  the  construc- 
tion, for  each  cable,  of  a  footbridge  or  working  platform  which 
permits  the  wires  to  be  observed  and  regulated  throughout  their 
length  and  greatly  facilitates  the  entire  work  on  the  cables. 

For  the  Manhattan  Bridge  (Fig.  46),  traveling  cages,  hanging 
from  the  footbridge  cables  as  a  track,  were  used  by  the  men 
placing  the  double  cantilever  floorbeams.  These  floorbeams, 
35!  feet  long  and  spaced  21  feet  apart,  were  supported  on  the 
footbridge  cables  in  pairs,  and  were  secured  to  the  upper  side  of 
these  cables  by  U-bolt  clamps.  Upon  the  outer  portions  of  the 
floorbeams  were  dapped  the  stringers,  three  lines  on  each  side, 
and  on  these  were  spiked  the  floorboards,  spaced  i|  inches 
apart  in  the  clear.  In  this  manner,  four  platforms  were  con- 
structed, 8  feet  wide,  placed  concentric  with  the  main  cables 
and  30  inches  (clear)  below  them.  The  platforms  were  provided 
with  wire-rope  handrails.  Passage  from  one  platform  to  another 
was  provided  only  at  the  towers  and  anchorages,  and  at  mid- 
span.  Each  platform  carried  nine  small  towers  called  "  hauling 
towers"  (Fig.  47),  about  250  feet  apart,  to  support  the  sheaves 
of  the  carrying  and  hauling  ropes  used  for  placing  the  strand 


168 


ERECTION  OF  SUSPENSION  BRIDGES 


wires.  The  platforms  were  braced  and  guyed  underneath  by 
backstays  from  each  tower,  and  by  inverted  storm  cables  con- 
nected to  them  at  intervals  of  54  feet.  The  entire  construction 
was  of  light  wooden  plank  (maximum  size  3X12)  and  all  con- 
nections were  thoroughly  bolted  with  washer  bearings.  All 
woodwork  had  been  previously  cut  to  length,  framed,  bored  and 
marked,  and  hoisted  to  the  tops  of  the  towers.  The  floorbeams 
were  slipped  down  on  the  cables  toward  the  center  of  the  span 


FIG.  47. — Manhattan  Bridge.     Footbridges  and  Sheave  Towers. 

and  toward  the  anchorages,  set  by  the  men  in  the  traveling 
cages,  and  maintained  in  position  by  the  stringers  dapped  to 
them.  Then  the  sheave  towers  and  handrails  were  erected  on 
the  platforms,  practically  completing  the  falsework  (Fig.  47). 

The  temporary  platforms  for  the  Williamsburg  Bridge  are 
shown  in  Fig.  57.  Two  footbridges  were  used,  67  feet  center  to 
center,  connected  by  transverse  bridges  every  160  feet. 

For  the  Brooklyn  Bridge,  Fig.  25,  the  timber  staging  con- 
sisted of  one  longitudinal  footbridge  and  five  transverse  plat- 


PARALLEL  WIRE  CABLES  169 

forms,  called  "  cradles,"  from  which  the  wires  were  handled  and 
regulated  during  cable-spinning. 

5.  Parallel  Wire  Cables. — Smaller  spans  have  been  built 
with  ready-made  parallel  wire  cables,  served  with  wire  wrapping 
at  short  intervals;   but  the  individual  wires  in  such  cables  lack 
freedom  to  adjust  themselves  to  the  necessary  curvature  of 
suspension,  so  that  objectionable  stress  conditions  arise.     For 
these  reasons  it  has  become  general  practice  to  use  the  method, 
introduced  by  Roebling,   of  spinning  the  desired  number  of 
parallel  wires  in  place  and  then  combining  them  into  a  cable. 
The  cable  is  pressed  into  cylindrical  form  and  wound   with 
continuous  wire  wrapping.     This  wrapping,  together  with  the 
tight  cable  bands  to  which  the  suspenders  are  attached,  serves 
to  create  enough  friction  pressure  between  the  wires  to  ensure 
united  stress  action. 

Guide  wires  are  used  as  a  means  of  adjusting  the  individual 
wires  to  equal  length.  Slight  differences  in  length,  if  distributed 
over  the  entire  span,  will  be  immaterial.  To  avoid  the  excess  in 
length  of  the  longer  wires  from  accumulating  at  a  single  point, 
the  wire  wrapping  should  be  started  at  a  considerable  number 
of  points  distributed  along  the  cable;  and  a  large  cable  should 
first  be  bound  into  smaller  temporary  strands  by  serving  with 
wire  at  intervals. 

The  length  of  the  guide  wire  must  be  accurately  computed, 
so  that  the  resulting  cable  shall  have  the  desired  sag  (assumed 
in  the  design)  after  the  bridge  is  completed.  Length  corrections 
must  be  made  for  any  cradling  of  the  cables,  variation  from  mean 
temperature,  the  curve  of  the  cable  saddle,  and  the  elastic 
elongation  due  to  the  suspended  load. 

6.  Initial  Erection  Adjustments. — Special  computations  have 
to  be  made  for  the  location  of  the  guide  wires,  for  setting  the 
saddles  on  top  of  the  towers,  and  for  the  length  of  the  strand  legs. 

When  the  desired  final  position  of  the  cable,  under  full  dead 
load,  is  known,  its  length  is  carefully  computed,  including  the 
main  span  parabola  between  points  of  tangency  at  the  saddles, 
the  short  curved  portions  in  the  saddles,  and  the  side-span 
parabolas  or  tangents  from  point  of  tangency  at  the  saddle  to 


170  ERECTION  OF  SUSPENSION  BRIDGES 

the  center  of  shoe  pin  at  the  anchorage.  Applying  corrections 
for  elastic  elongation  (due  to  suspended  load)  and  for  difference 
of  temperature  from  the  assumed  mean,  the  length  of  unloaded 
cable  is  determined.  This  gives  the  length  of  the  guide  wire 
between  the  same  points. 

Assuming  no  slipping  of  the  strands  in  the  saddles,  the 
initial  position  of  the  saddles  is  computed  so  as  to  balance 
tensions  (or  values  of  y)  between  the  main  and  side  span  cate- 
naries. This  gives  the  distance  the  saddles  must  be  set  back 
(toward  shore)  from  their  final  position  on  the  tops  of  the  towers. 

Since  the  strands  will  be  spun  about  2  feet  above  their  final 
position  in  the  tower  saddles,  the  initial  position  of  the  strand 
shoes  will  be  a  short  distance  forward  of  their  final  position. 
This  distance  is  carefully  computed  and  gives  the  required 
length  of  the  "  strand  legs"  (Fig.  48).  The  distance  may  also  be 
determined  or  checked  by  actual  trial  with  the  guide  wire. 

Taking  into  consideration  the  previously  calculated  and  cor- 
rected total  length  of  cable  between  strand  shoes,  the  initial 
raised  position  of  the  strands  above  the  tower  saddles,  and  the 
length  of  strand  legs  shifting  the  initial  position  of  the  strand 
shoes,  the  ordinates  of  the  initial  catenaries  in  main  and  side 
spans  are  carefully  computed.  These  ordinates  are  used  for 
setting  the  guide  wires  with  the  aid  of  level  and  transit  stationed 
at  towers  and  anchorages. 

For  the  Cumberland  River  footbridge  (540-foot  span,  see  page 
184),  the  saddles  on  the  two  towers  were  set  back  about  5  inches 
toward  shore  from  center  of  tower.  This  distance  was  figured 
from  backstay  elongation  and  tower  shortening  due  to  dead  load 
plus  one-half  live  load,  so  that  the  center  of  the  tower  would 
bisect  the  movement  of  the  shoe  (on  rollers)  for  live  load  at 
mean  temperature.  Allowing  for  displacement  of  saddles  and 
cable  stretch,  the  no-load  cable-sag  was  made  38^  feet  in  order  that 
the  sag  in  final  position  under  full  live  load  should  be  45  feet. 

In  the  case  of  the  Brooklyn  Bridge,  the  strands  were  spun 
about  57  feet  above  their  final  position  at  mid-span,  the  purpose, 
as  stated,  being  to  avoid  interference  with  regulation  and  to 
increase  the  tension  as  an  initial  strength-test  of  the  individual 


INITIAL  ERECTION  ADJUSTMENTS 


171 


wires.  In  consequence,  the  strand  leg  had  to  be  designed  so  as 
to  hold  the  shoe  12  feet  behind  the  anchor  pin.  After  the  strand 
was  finished,  the  shoes  were  let  forward  into  their  final  places 
and,  at  the  same  time,  the  strand  was  lowered  from  the  rollers 
on  top  of  the  saddle  into  the  saddle,  which  double  operation 
caused  the  vertex  to  sink  into  correct  position  as  previously 
calculated. 

For  the  Williamsburg  Bridge,  the  strands  were  spun  15  feet 
above  their  final  position,  requiring  the  shoes  to  be  initially  set 
back  of  the  anchor  pin,  as  in  the  Brooklyn  Bridge. 

For  the  Manhattan  and  Kingston  Bridges  (H.  D.  Robinson, 
Engineer-in-charge) ,  the  strands  were  spun  parallel  to  (and 
slightly  above)  their  final  position.  In  these  cases,  the  strand 
leg  held  the  shoe  a  short  distance  in  front  of  the  anchor  pin; 
and  the  shoe  had  to  be  pulled  back  that  distance  when  the 
strands  were  lowered  into  the  saddle  (Fig.  49). 

In  the  case  of  the  Kingston  Bridge  (yo5-foot  span,  Fig.  56), 
instead  of  setting  the  saddles  back  on  the  towers,  the  tops  of  the 
towers  were  tipped  back  toward  the  shore  a  distance  of  6  inches, 
by  means  of  temporary  backstays,  the  anchor  bolts  at  the  toe 
being  loosened  J  inch  to  permit  the  tilting.  The  cables  were 
erected  with  the  towers  and  the  attached  saddles  in  this  position. 
As  the  steelwork  in  the  main  span  was  erected,  the  backstays 
gradually  elongated  until  the  towers  returned  to  their  final 
vertical  position. 

The  initial  erection  adjustments  for  the  Brooklyn,  Williams- 
burg,  Manhattan  and  Kingston  Bridges  are  summarized  arid 
compared  in  the  following  table: 

INITIAL  POSITION  OF  CABLE  STRANDS 
(With  Reference  to  Final  Position) 


Height 

Height 

Distance 

Distance 

Above 

Above 

Saddle 

Shoe 

Crown 

Saddle 

Set  Back 

Set  Forward 

Brooklyn  

57  ft- 

2  .  I  ft. 

O.I   ft. 

-12  ft. 

Williamsburg  

15 

2 

2-75 

-  3 

Manhattan  

2 

2 

0 

+  1-83 

Kinjrs^on  

1-25 

1-25 

o-5 

+    0.2$ 

172 


ERECTION  OF  SUSPENSION  BRIDGES 


7.  Spinning  of  Cables. — The  operation  of  cable-spinning 
requires  an  endless  wire  rope  or  "  traveling  rope"  (Fig.  48)  sus- 
pended across  the  river  and  driven  back  and  forth  by  machinery 


FIG.  48. — Strand  Shoes  and  Traveling  Sheaves  Ready  for  Cable  Spinning. 
(Manhattan  Bridge.) 

for  the  purpose  of  drawing  the  individual  wires  for  the  cable 
from  one  anchorage  to  the  other.  There  is  also  suspended  a 
"guide  wire"  which  is  established  by  computations  and  re'gu- 


CABLE  SPINNING  173 

lated  by  instrumental  observations  so  as  to  give  the  desired 
deflection  of  the  cable  wires. 

Large  reels,  upon  which  the  wires  are  wound,  are  placed  at 
the  ends  of  the  bridge  alongside  the  anchor  chains  (Fig.  48) .  The 
free  end  of  a  wire  is  fastened  around  a  grooved  casting  of  horse- 
shoe outline  called  a  "shoe"  (Fig.  48),  and  the  loop  or  bight, 
thus  formed  is  hung  around  a  light  grooved  wheel  (Fig.  48)  which 
is  fastened  to  the  traveling  rope.  The  traveling  rope  with  its 
attached  wheel,  moving  toward  the  other  end  of  the  bridge, 
thus  draws  two  parts  of  the  wire  simultaneously  across  from 
one  anchorage  to  the  other;  one  of  these  parts,  having  its  end 
fixed  to  the  shoe,  is  called  the  "standing  wire";  while  the  other, 
having  its  end  on  the  reel,  is  called  the  "running  wire"  and 
moves  forward  with  twice  the  speed  of  the  traveling  rope. 
Arriving  at  the  other  end,  the  wire  loop  is  taken  off  the  wheel 
and  laid  around  the  shoe  at  that  end.  The  two  parts  of  the 
wire  are  then  adjusted  so  as  to  be  accurately  parallel  to  the  guide 
wire,  the  operation  of  adjustment  being  controlled  by  signals 
from  men  stationed  along  the  footbridge.  The  wire  is  then 
temporarily  secured  around  the  shoe,  and  a  new  loop  hung  on 
the  traveling  wheel  for  its  second  trip.  After  two  or  three 
hundred  wires  have  thus  been  drawn  across  the  river  and  accu- 
rately set,  they  are  tied  together  at  intervals  to  form  a  cable 
strand. 

For  the  Manhattan  Bridge,  the  wires  (drawn  in  3ooo-foot 
lengths)  were  spliced  to  make  a  continuous  length  of  80,000 
feet  (4  tons)  wound  on  a  wooden  reel  (Fig.  48).  These  reels 
were  48  inches  in  diameter  (at  bottom  of  groove)  and  26  inches 
long,  and  were  provided  with  brake  drums.  On  each  anchorage 
were  set  eight  reel  stands,  each  with  a  capacity  of  four  reels. 

The  equipment  used  for  cable-spinning  consisted  of  an  end- 
less f-inch  steel  traveling  rope  passing  around  a  6-foot  hori- 
zontal sheave  at  each  anchorage;  machinery  for  operating  the 
endless  rope;  devices  for  removing  and  adjusting  the  wires  and 
strands;  apparatus  for  compacting  and  wrapping  the  cables; 
hoisting  machinery  and  power  plant. 

Attached  to  the  endless  rope  (" traveling  rope")   at  two 


174 


ERECTION  OF  SUSPENSION  BRIDGES 


equidistant  points,  were  deeply-grooved  4-foot  carrier  sheaves 
(" traveling  wheels")  in  goose-neck  frames  (Fig.  48).  These 
frames  were  held  securely  in  a  vertical  plane,  and  were  designed 
with  clearance  to  ride  over  the  supporting  sheaves. 

The  "strand  shoe"  was  held  22  inches  in  front  of  final  posi- 
tion by  a  special  steel  construction  called  a  " strand  leg"  (Fig.  48) 
attached  to  the  pin  between  two  anchorage  eyebars. 

The  bight  of  wire  was  placed  around  the  traveling  wheel  and 


FIG.  49. — Manhattan  Bridge.     Anchoring  a  Completed  Strand. 

pulled  across.  As  each  part  of  the  wire  became  dead,  it  was 
taken  by  an  automatic  Buffalo  grip  at  the  tower  and,  with  a 
4-part  handtackle  of  manila  rope,  adjusted  to  the  guide  wire. 
It  required  about  seven  minutes  for  a  trip  across  from  anchorage 
to  anchorage  (3223  feet).  Only  ten  field  splices  were  required 
to  a  strand  (256  wires).  After  the  strand  was  completed,  the 
wires  were  compacted  with  curved- jaw  tongs  and  fastened  (or 
"seized")  with  a  few  turns  of  wire,  every  10  feet.  Then,  with  a 
"strand-bridle"  attached  to  a  35-ton  hydraulic  jack  (Fig.  49), 


CABLE  SPINNING  175 

the  shoe  was  pulled  toward  the  shore,  releasing  the  strand  leg 
and  the  eyebar  pin.  The  strand  shoe  was  then  revolved  90° 
to  a  vertical  position  (Fig.  49),  and  pulled  back  to  position  on 
the  eyebar  pin. 

The  strand  was  then  lifted  from  the  temporary  sheaves  in 
which  it  was  laid  at  the  anchorages  and  the  towers,  and  lowered 
into  the  permanent  saddles;  a  20-ton  chain  hoist  and  steel 
"balance  beam"  were  used  for  this  operation.  The  strand  was 
then  adjusted  to  the  exact  deflection  desired,  by  means  of  shims 
in  the  strand  shoe. 

After  the  seven  center  strands  were  completed,  they  were 
bunched  together  with  powerful  squeezers  to  make  a  cylinder 
about  9!  inches  in  diameter,  secured  with  wire  " seizings"  at 
intervals.  Then  the  remaining  strands  were  completed,  and 
compacted  in  two  successive  layers  around  the  core,  the  inter- 
stices being  filled  with  petrolatum.  A  hydraulic  compacting 
machine  was  used  for  this  squeezing,  and  temporary  clamps 
applied. 

Then  the  cable  was  coated  with  red-lead  paste,  and  the 
permanent  cable  bands  and  suspenders  were  attached. 

After  the  stiffening  trusses  and  floor  were  suspended,  the 
spaces  between  the  cable  bands  were  covered  with  wire  wrap- 
ping. 

The  spinning  of  these  cables  took  six  days  for  a  strand  (256 
wires) ;  but  four  strands  in  each  cable  were  strung  simultaneously. 
The  four  cables  (each  consisting  of  thirty-seven  strands  or  9472 
wires)  were  completed  in  four  months.  The  work  of  compres- 
sing and  binding  the  cables  and  attaching  the  suspender  clamps 
and  ropes  took  two  or  three  months  more,  but  the  erection  of  the 
suspended  trusses  proceeded  at  the  same  time. 

As  soon  as  the  .strands  were  completed,  the  footbridges  were 
hung  to  the  main  cables  to  be  later  used  for  the  work  of  cable 
wrapping.  The  temporary  footbridge  cables  were  cut  up  for 
use  as  suspenders. 

For  the  Williamsburg  Bridge  (Fig.  57),  the  wire  was  supplied 
on  7-foot  wooden  reels  carrying  90,000  feet  (9000  pounds)  per 
reel.  An  engine  on  the  New  York  s'de  operated  the  driving 


176  ERECTION  OF  SUSPENSION  BRIDGES 

wheels  around  which  two  endless  ropes  passed.  Two  carrier 
sheaves  on  each  endless  rope  traveled  back  and  forth,  carrying 
two  bights  across  (for  two  strands)  on  the  forward  trip,  and  two 
bights  (for  two  adjacent  strands)  on  the  return  trip.  In  this 
manner  each  endless  rope  was  laying  four  strands  at  the  rate  of 
fifty  wires  in  each  strand  in  ten  hours. 

Eight  reels  of  wire  were  required  for  each  strand.  When  the 
end  of  a  coil  was  reached,  it  was  held  in  a  vise  and  connected  to  a 
wire  from  a  fresh  reel,  by  screwing  up  a  sleeve  nut  over  the  screw- 
threaded  ends  (which  were  formed  by  a  special  machine  to  roll 
the  threads). 

As  the  wire  was  laid,  it  was  adjusted  to  conform  to  the 
catenary  of  the  guide  wire,  in  order  to  secure  uniform  tension 
in  the  wires  of  the  finished  cable. 

The  carrier  wheels  moved  400  feet  per  minute.  There  were 
three  men  at  each  anchorage  to  handle  the  reels,  make  splices, 
adjust  the  wire  and  take  the  bights  off  and  on  the  carrier  wheels. 
As  the  carrier  wheel  passed  each  tower,  three  men  on  the  top  of 
the  tower  clamped  handtackle  to  the  wire  and  pulled  up  until 
the  wire  was  adjusted  exactly  parallel  to  the  guide  wire,  as 
signaled  by  men  distributed  along  the  footbridge  (three  men  on 
each  side  span  and  seven  men  on  the  main  span).  These  men 
clamped  the  wire  to  the  strand  after  adjustment.  After  the 
standing  wire  was  adjusted,  the  running  wire  was  regulated  in 
the  same  manner,  but  in  the  reverse  order.  A  total  of  twenty- 
five  men  were  thus  required  to  handle  the  wire  as  it  was  laid. 

As  soon  as  the  strand  was  completed,  the  shoe  was  drawn 
clear  of  its  support  by  a  25-ton  ratchet  jack  anchored  to  the 
masonry.  Then  the  shoe  was  twisted  by  hand  with  a  long  bar 
and  thus  revolved  90°  into  a  vertical  plane  and  allowed  to  slip 
back  towards  the  tower,  thus  lowering  the  strand  in  the  middle 
of  the  main  span.  The  shoe  was  then  permanently  connected 
to  the  end  pin  of  the  anchor-chain  eyebars.  Shims  back  of  the 
pin  in  the  slotted  pin-hole  of  the  strand  shoe  provided  adjust- 
ment for  the  strand  length;  each  f-inch  shim  corresponded  to  a 
vertical  movement  of  about  i  inch  at  mid-span. 

When  the  inner  strands  were  completed,   their  ties  were 


COMPACTING  THE  CABLES  177 

removed  and  they  were  made  into  one  strand  to  avoid  trouble 
in  handling  them  after  they  were  surrounded  by  the  remaining 
strands. 

8.  Compacting  the  Cables. — Each  cable  consists  of  3,  7,  19 
or  37  strands,  depending  upon  its  size,  and  these  have  to  be 
compacted  to  make  a  cylindrical  cable. 

For  the  Manhattan  Bridge,  the  temporary  seizings  around 
the  strands  were  removed  and  the  cable  was  compacted  by 
hydraulic  squeezers.  Sixteen  duplicate  squeezers  were  used, 
each  consisting  of  a  hinged  collar  with  a  hydraulic  jack  of  6-inch 
stroke  opposite  the  hinge.  A  hydraulic  hand-pressure  pump 
was  used  to  produce  a  pressure  of  5000  pounds  per  square  inch 
or  a  total  force  of  43,000  pounds  on  the  squeezer  piston.  Seizing 
(12  turns  of  No.  8  wire)  was  applied  close  to  the  squeezer,  which 
was  then  moved  2  feet  forward  to  repeat  the  operation.  With 
two  men  operating  each  squeezer,  the  four  cables  were  com- 
pacted in  a  few  weeks. 

9.  Placing  Cable  Bands  and  Suspenders. — After  the  cables 
are  compacted  (with  wire  seizing  at  short  intervals  to  hold  them) , 
the  cable  bands  are  placed  at  the  panel  points. 

For  the  Manhattan  Bridge,  the  cable  bands  (Fig.  55)  con- 
sist of  split  cast-steel  sleeves,  3  feet  long,  with  ten  if -inch  bolts 
through  the  longitudinal  flanges.  The  upper  half  has  two 
semicircular  grooves,  12  inches  apart,  for  holding  the  suspender 
ropes.  The  bolts  were  screwed  up  tight  by  means  of  socket 
wrenches  with  4-foot  handles,  operated  by  two  or  three  men 
each. 

The  if -inch  suspender  ropes,  made  by  cutting  up  the  tem- 
porary footbridge  cables,  were  fitted  with  cast-steel  sockets 
5!  inches  in  diameter  by  17  inches  long.  These  sockets  were 
threaded  on  the  outside  to  receive  a  cast-steel  nut  5^  inches 
thick.  The  ends  of  the  rope  were  served;  and  the  wires  beyond 
the  serving  were  spread,  cleaned  in  dilute  acid,  washed  in  water 
and  dried  with  a  painter's  torch.  The  end  of  the  rope  was  then 
passed  through  the  socket  (which  had  been  carefully  cleaned  of 
sand  and  scale) ,  the  wires  were  spread  to  fill  the  covered  portion, 
and  melted  spelter  (heated  to  a  very  thin  consistency)  was 


178 


ERECTION  OF  SUSPENSION  BRIDGES 


poured  in,  filling  all  the  interstices.  Some  of  the  finished  ropes 
were  tested,  and  showed  an  ultimate  strength  of  287,000  to 
290,000  pounds,  with  the  rope  breaking  4  to  8  feet  from  the 
socket;  there  was  no  sign  of  injury  at  the  socket,  thread  or 
nut. 

The  suspenders  were  then  placed  in  position  around  the 
cable  bands,  with  their  lower  ends  ready  to  engage  the  bottom 
chords  of  the  stiffening  trusses  (Fig.  50). 


FIG.   50. — Manhattan  Bridge.     Erection  of  Lower  Chords  and  Floor  System. 

10.  Erection  of  Trusses  and  Floor  System. — The  suspension 
from  the  cables  permits  the  steelwork  to  be  erected  without 
falsework.  In  planning  the  program  of  erection  there  must  be 
considered  the  method  of  connection  to  the  suspenders,  clear- 
ances for  travelers,  and  the  reach  of  the  booms.  In  addition, 
the  scheme  should  aim  to  balance  the  dead-load  distribution 
along  the  span,  so  as  to  minimize  the  distortion  of  the  cables 
during  erection. 

In  the  Manhattan  Bridge,  the  truss  is  supported  at  each 


ERECTION  OF  TRUSSES  AND   FLOOR  SYSTEM 


179 


panel  point  by  four  parts  of  if -inch  steel  rope  suspenders  (Fig.  50) 
with  their  bights  engaging  the  main  cables  and  having,  at  the 
lower  end,  nut  bearings  on  horizontal  plates  across  the  bottom 
flanges  of  the  lower  chord. 

All  members  were  shipped  separately,  the  chord  members  in 
two-panel-length  pieces  weighing  26,000  to  30,000  pounds  each. 

The  erection  proceeded  at  four  points  simultaneously,  work- 
ing in  both  directions  from  each  tower  (Fig.  50) .  Traveler  der- 


FIG.  51. — Manhattan  Bridge.     Erection  of  Verticals. 

ricks  of  25-ton  capacity  were  used,  with  34-foot  mast  and  5o-foot 
boom  (covering  two  panels  in  advance)  and  provided  with  bull- 
wheel.  At  each  point  of  erection  there  were  two  of  these  large 
derricks,  also  one  jinnywink  derrick  with  3O-foot  boom  and 
7- ton  capacity.  In  addition  to  these  twelve  movable  derricks, 
there  were  four  stationary  steel-boom  derricks  at  the  towers. 

Starting  at  the  towers,  the  lower  chords  and  floor  system 
were  assembled  two  panels  in  advance  of  the  travelers,  making 
temporary  connections  to  the  suspenders,  until  the  anchorages 


180 


ERECTION  OF  SUSPENSION  BRIDGES 


and  mid-span  were  reached.     Then  the  travelers  returned  to  the 
towers  to  commence  their  second  trip. 

The  material  was  hoisted  by  the  tower  derricks  and  loaded 
on  service  cars  which  delivered  it  to  the  traveler  derricks.  The 
service  cars  ran  on  the  permanent  track  between  the  inner  and 
outer  trusses;  the  cars  were  hauled  away  from  the  tower  by 
cables  operated  by  hoisting  engines  on  the  tower,  and  returned 
empty,  by  gravity,  on  the  grade  furnished  by  the  camber. 


FIG.  52. — Manhattan  Bridge.    Erection  of  Diagonals. 

On  the  first  trip,  the  lower  chords,  lower  deck  and  verticals 
were  erected  (Fig.  51);  on  the  second  trip,  the  truss  diagonals 
were  erected  (Figs.  52,  53);  and  on  the  return  (Fig.  54),  the  upper 
deck  and  transverse  bracing  were  put  up,  thus  completing  the 
structure. 

On  the  first  trip,  temporary  suspender  connections  were  made 
to  the  lower  chord  at  alternate  panel  points  so  as  to  miss  the 
upper  chord  splices.  After  the  return  of  the  travelers,  permanent 
connection  and  adjustment  of  the  suspenders  at  the  other  points 


ERECTION  OF  TRUSSES  AND   FLOOR   SYSTEM 


181 


were  made.  The  temporary  suspender  connections  were  removed 
before  the  top  chords  were  erected,  and  were  connected  again 
(permanently)  after  the  top  chords  were  in  place. 

A  force  of  three  hundred  men  was  employed  on  this  work, 
and  their  record  was  300  tons  of  steel  erected  in  a  day.  There 
were  about  1,000,000  field  rivets  in  the  three  spans.  The  bridge 
was  formally  opened  ten  months  after  the  floor  hanging  com- 
menced. 

Where  the  side  spans  are  not  suspended  from  the  cables, 


FIG.  53.— Manhattan  Bridge,    View  before  Erection  of  Top  Chords. 

falsework  is  generally  required.  In  the  Kingston  Suspension 
Bridge,  Fig.  56,  the  side  spans  (although  suspended)  were 
erected  on  light  falsework,  as  time  was  thereby  saved. 

The  first  few  panels  of  the  main  span  are  generally  erected 
by  the  stationary  derricks  at  the  tower,  as  far  as  their  booms 
can  reach.  Additional  panels  may  be  erected  by  drifting  or 
outhauling  from  the  cable;  or  by  the  use  of  " runners,"  that  is, 
block  and  falls  suspended  from  the  advance  cable  band  and 


182 


ERECTION  OF  SUSPENSION  BRIDGES 


operated  by  the  hoisting  engine  at  the  tower.  At  Kingston,  the 
latter  method  was  adopted,  dispensing  with  the  use  of  travelers 
for  the  main  portion  of  the  span. 

11.  Final  Erection  Adjustments. — The  equilibrium  polygon 
is  computed  for  the  dead  load  acting  on  the  cable,  and  levels 
taken  at  a  number  of  points  on  the  cable  should  check  these 
ordinates.  The  elevations  and  camber  of  the  roadway  are  also 


FIG.  54. — Manhattan  Bridge.     Erection  of  Top  Chords. 

checked  with  levels  and  corrected,  where  necessary,  by  adjusting 
the  lengths  of  the  suspenders. 

In  completing  the  stiffening  truss,  the  closing  chord  members 
should  be  inserted  after  all  the  dead  load  is  on  the  structure,  the 
connecting  holes  at  one  end  being  drilled  in  the  field. 

If  the  closure  of  the  stiffening  truss  has  to  be  made  before 
full  dead  load  is  on  the  structure,  or  at  other  than  mean  tempera- 
ture, the  vertical  deflections  are  computed  for  these  variations 
from  assumed  normal  conditions  and  the  suspenders  adjusted 
accordingly,  before  connecting  the  closing  members. 


CABLE  WRAPPING  183 

In  adjusting  the  suspenders,  the  center  hanger  is  shortened 
or  lengthened  the  calculated  amount,  and  the  other  hangers  are 
corrected  by  amounts  varying  as  the  ordinates  to  a  parabola. 

If  the  trusses  are  assembled  on  the  ground  before  erection, 
the  exact  camber  ordinates  can  be  measured  and  reproduced  (by 
suspender  adjustment),  so  as  to  secure  zero  stress  under  full 
dead  load  at  mean  temperature. 

An  ideal  method  of  checking  the  final  adjustments  is  by 
means  of  an  extensometer,  which  should  check  zero  stresses 
throughout  the  stiffening  truss  when  normal  conditions  are 
attained,  or  calculated  stresses  for  any  variation  from  assumed 
normal  conditions. 

Instead  of  adjusting  to  zero  stress  for  full  dead  load,  it  would 
be  more  scientific  and  somewhat  more  economical  to  adjust  for 
zero  stress  at  dead  load  plus  one-half  live  load. 

12.  Cable  Wrapping. — Close  wire  wrapping  has  proved  to  be 
the  most  effective  protection  for  cables. 

For  the  Manhattan  Bridge,  No.  9  galvanized  soft-steel  wire 
(o.  148-inch  diameter)  was  used.  This  was  rapidly  wound 
around  "the  cable  by  a  very  simple  and  ingenious  self-propelling 
machine  operated  by  an  electric  motor.  This  machine,  designed 
by  H.  D.  Robinson,  is  illustrated  in  Fig.  55. 

In  advance  of  the  machine,  the  temporary  seizings  are  care- 
fully removed  and  the  cable  painted  with  a  stiff  coat  of  red- 
lead  paste.  The  end  of  the  wrapping  wire  is  fastened  in  a  hole 
in  the  groove  at  the  end  of  the  cable  band.  The  machine,  carry- 
ing the  wire  on  two  bobbins  or  spools,  travels  around  the  cable 
and  applies  the  wire  under  a  constant  tension.  The  machine 
presses  the  wire  against  the  preceding  coil  and  at  the  same  time 
pushes  itself  along.  The  rate  is  about  18  feet  per  hour. 

The  machine  weighs  1000  pounds  and  is  operated  by  a 
i|-H.P.  motor  at  a  speed  of  13  R.P.M.  It  is  handled  by  a  force 
of  six  men. 

(The  small  hand-operated  device,  which  was  superseded  by 
the  motor-driven  machine,  is  seen  at  the  extreme  right  in 
Fig.  55.  It  was  used  to  complete  the  wrapping  close  to  the 
cable  bands.) 


184 


ERECTION  OF  SUSPENSION  BRIDGES 


For  the  Williamsburg  Bridge,  wire  wrapping  was  not  used; 
instead,  the  cables  were  covered  with  a  preservative  coating  of 
oil  and  graphite,  then  wrapped  spirally  with  three  layers  of 
waterproof  duck,  and  finally  enclosed  in  a  thin  steel-plate  shell 
made  in  two  semi-cylindrical  portions  with  overlapping  joints 
and  locked  fastenings.  This  protection  has  proved  inadequate 
to  keep  out  moisture  and  prevent  rust,  and  it  has  recently  (1917- 
1921)  been  replaced  by  wire  wrapping  applied  with  Robinson's 
machine. 


FIG.  55. — Manhattan  Bridge.    Cable  Wrapping  Machine. 

13.  Erection  of  Wire  Rope  Cables. — The  individual  wire 
ropes  composing  a  cable  of  this  type  may  be  towed  across  the 
river  in  the  same  manner  as  the  temporary  footbridge  ropes  of  a 
parallel  wire  cable;  or  they  may  be  strung  across  by  means  of  a 
single  working  cable  stretched  from  tower  to  tower. 

The  latter  method  was  used  for  a  footbridge  of  54o-foot  span 
built  in  1919  over  the  Cumberland  River  by  the  American  Bridge 
Co.  Each  cable  consisted  of  seven  ropes  of  if -inch  diameter. 


ERECTION  OF  WIRE  ROPE  CABLES 


185 


A  working  cable  of  i-inch  wire  rope  was  first  stretched  across 
between  the  towers  for  each  of  the  main  cables.  The  main 
ropes  were  unwound  from  the  reels  back  of  one  tower.  One 
end  of  a  rope  was  lifted  to  the  top  of  the  tower  and  hauled 
across  the  river  to  the  top  of  the  opposite  tower,  the  rope  being 
supported  from  the  i-inch  working  cable  by  blocks  attached  at 
intervals  of  about  60  feet,  thus  preventing  too  much  sag.  The 


FIG.  56. — Erection  of  Rondout  Creek  Bridge  at  Kingston,  N.  Y.,  1921. 

Type  OS. 

Span  705  feet. 

rope  was  then  lowered  to  approximately  correct  position,  and 
the  sockets  attached  to  the  tower  shoes.  The  remaining  ropes 
were  then  stretched  in  the  same  manner,  and  all  were  then 
adjusted  by  nuts  at  the  ends  until  they  touched  a  level  straight- 
edge held  on  the  fixed  line  of  sag  determined  by  a  transit  in  the 
tower.  The  cable  clamps  and  suspenders  were  then  placed  by 
men  on  a  movable  working  platform  hung  from  the  cables, 
beginning  in  the  center  and  working  toward  each  end.  A 
"boatswain  chair"  was  used  to  carry  out  men,  materials  and 


186 


ERECTION  OF  SUSPENSION  BRIDGES 


tools.  The  floor  system  was  also  erected  by  men  on  the  work- 
ing platform,  in  this  case  working  from  both  ends  toward  the 
center.  The  platform  was  then  removed,  and  the  trusses  were 
erected  from  the  ends  toward  the  center  by  workmen  on  the 
floor  system,  using  the  two  working  cables  (shifted  to  the  center 
of  the  bridge)  as  a  trolley  cable  for  transporting  the  truss  sec- 
tions to  position.  When  the  top  lateral  bracing,  railings,  and 
wood  floor  were  added,  the  structure  was  completed.  A  total 


FIG.  57. — Footbridges  for  Erection  of  Williamsburg  Bridge. 
(See  Fig.  31,  page  88). 

of  205  tons  of  structural  steel  and  45  tons  of  cables  were  thus 
erected  in  a  period  of  twelve  weeks. 

14.  Erection  of  Eyebar  Chain  Bridges. — Chain  suspension 
bridges  have,  as  a  rule,  been  erected  upon  falsework. 

The  falsework  used  for  the  erection  of  the  Elizabeth  Bridge 
at  Budapest  (1902,  Span  951  feet)  is  shown  in  Fig.  58.  The 
falsework  consisted  of  huge  scaffoldings  built  on  piles  and 
protected  from  floating  ice  by  ice  breakers.  Four  openings  of 
1 60  feet  were  left  for  vessels;  these  openings  were  spanned  by 


ERECTION  OF  EYEBAR  CHAIN  BRIDGES 


187 


temporary  timber  bridges  floated  into  place  on  pontoons.  After 
the  falsework  was  completed,  the  main  chains  were  erected  in 
twelve  weeks.  The  falsework  was  then  taken  down  and  the 
steelwork  completed. 

At  a  crossing  like  the  East  River  or  the  Hudson  River,  the 
use  of  such  falsework  would  be  out  of  the  question.  A  com- 
parison of  the  cumbersome  construction  employed  for  the  Eliza- 
beth Bridge  (Fig.  58)  with  the  comparatively  insignificant 
scaffolding  required  for  the  Williamsburg  Bridge  (Fig.  57),  is 
an  argument  for  wire  cable  vs.  eyebar  bridges. 


FIG.  58.— Falsework  for  the  Elizabeth  Bridge  (Eyebar  Chains). 
(See  Fig.  34,  page  95). 

A  different  scheme,  eliminating  heavy  falsework,  was  used 
for  the  Clifton  Bridge  (1864,  Span  702  feet).  Under  each  set  of 
three  chains,  a  suspension  footbridge  was  constructed,  using 
wire  ropes.  Above  this  staging,  another  rope  was  suspended 
to  carry  the  trolley  frames  for  transporting  the  links.  The 
chains  were  commenced  simultaneously  at  the  two  anchor  plates, 
the  lowest  of  the  three  chains  being  put  in  first.  Commencing 
at  the  anchorage,  there  were  inserted  the  whole  number  of  links, 
namely  12,  then  n,  10,  9,  8,  and  so  on  until  the  chain  was 


188  ERECTION  OF  SUSPENSION  BRIDGES 

diminished  to  i  link;  then  the  chain  was  continued  with  i  and 
2  links,  alternately,  until  the  two  halves  met  at  mid-span.  The 
suspended  footbridge  was  strong  enough  to  carry  the  weight  of 
this  chain  (consisting  of  i  and  2  links,  alternately)  until  the 
center  connection  was  made;  the  chain  was  then  made  to  take 
its  own  weight  by  removing  the  blocking  under  it.  The  next 
operation  was  to  add  the  remaining  links  of  the  chain  on  the 
pins  already  in  place.  The  process  was  repeated  for  the  upper 
chains,  and  then  the  roadway  was  suspended. 

The  Cologne  Suspension  Bridge  (1915,  Span  605  feet, 
Fig.  17),  was  the  first  large  bridge  to  be  built  hingeless.  (The 
Kingston  Bridge,  1921,  was  the  second.)  It  is  of  the  self- 
anchored  type,  the  stiffening  girder  taking  up  the  horizontal 
tension;  and  the  towers  are  hinged  at  the  base.  Nickel  steel 
was  used  for  the  chains,  the  eyebars  being  of  the  European  type, 
that  is,  of  flat  plates  (36  to  59  inches  wide)  bored  for  1 2-inch 
pin-holes  near  the  ends.  The  erection  of  the  chains  and  stiffen- 
ing girders  proceeded  simultaneously  on  special  staging,  and 
was  so  conducted  that  the  girders  were  completed  first.  The 
girders  were  made  three-hinged  during  erection  and  then  changed 
to  hingeless  by  riveting  on  splice  plates. 

The  procedure  was  as  follows:  Falsework  was  built  for  the 
side  spans  and  a  traveler  was  assembled  at  each  end.  The  side- 
span  girders  and  deck  were  erected  on  the  falsework,  and  the 
staging  built  up  (on  the  girders)  for  the  land  chains,  the  traveler 
moving  forward  from  the  anchorage  to  the  tower  during  this 
operation.  The  traveler  then  moved  out  on  cantilever  false- 
work spans  in  the  main  opening,  erecting  the  stiffening  girders 
and  the  staging  for  the  chains  from  the  tower  to  mid-span. 
The  erection  of  the  chains  followed  closely  upon  the  erection  of 
the  girders.  When  the  stiffening  girders  were  completed  and 
the  suspension  chains  connected  to  the  ends  (with  24-inch  pins), 
every  third  hanger  was  coupled  up.  The  staging  carrying  the 
chains  was  then  removed,  and  the  remaining  hangers  were  con- 
nected and  adjusted  by  means  of  their  turnbuckles  to  bring  the 
pin  points  in  the  chains  into  their  correct  positions.  The  splices 
in  the  webs  and  flanges  of  the  stiffening  girders  at  the 


TIME  REQUIRED   FOR  ERECTION  189 

three  hinge-points  were  then  riveted  up,  thus  completing  the 
erection. 

For  LindenthaFs  Quebec  Design  (1910,  Span  1758  feet,  Fig. 
40),  the  following  scheme  of  erection  was  proposed:  The  side 
spans  were  to  be  erected  on  steel  falsework — first  the  floor  system, 
then  the  eyebars  and  pins  of  the  lower  chord  chain,  then  the 
verticals  and  upper  chain  eyebars,  leaving  the  pins  projecting 
out  to  receive  the  diagonals  and  remaining  eyebars  after  the 
main  span  chains  were  erected  and  self-supporting.  The 
towers  were  to  be  riveted  up  in  place  and  temporarily  anchored 
to  the  steel  staging  which,  in  turn,  was  to  be  anchored  to  the 
abutment.  The  first  sets  of  eyebars  (one  and  two  alternating 
per  panel)  of  the  chains  of  the  middle  span  were  to  be  erected 
from  temporary  wire  rope  cables,  each  consisting  of  forty  steel 
wire  ropes  of  2^-inch  diameter.  Then  the  remaining  eyebars 
and  gusset  plates  were  to  be  pushed  on  to  the  pins  until  the 
chains  were  completed.  Thereafter  the  verticals  and  diagonals 
were  slipped  in  place,  and  the  suspenders  and  floor  system  of  the 
middle  span  erected. 

15.  Time  Required  for  Erection. — The  time  schedule  for  the 
Manhattan  Bridge  (i47o-foot  span,  Fig.  35)  was  as  follows: 

First  substructure  contracts  let 1901 

Pier  foundations  commenced May,  1901 

Work  commenced  on  final  (revised)  design March,          1904 

Steel  towers  commenced July,  1907 

Steel  towers  completed  (12,500  tons) July>  1908 

Temporary  cables  strung June  15-20,  1908 

Footbridges  constructed July    7~*3>  1908 

Spinning  of  main  cables  commenced  (4  cables) .  .  .   Aug.  10,        1908 

Last  wire  strung  (37,888  wires) :   Dec.  10,        1908 

Erection  of  suspended  steel  commenced Feb.  23,       1909 

Erection  of  suspended  steel  completed  (24,000  tons)  June     i,        1909 
Approaches  completed  and  bridge  formally  opened.  Dec.  31,        1909 

The  steel  erection,  amounting  to  42,000  tons  of  steel  between 
anchorages  and  including  towers,  cables,  trusses  and  decks,  was 
accomplished  in  two  and  a  half  years. 

The  Kingston  Suspension  Bridge  (705-foot  span,  Fig.  56) 
was  completed  in  one  year  (1920-1921),  although  several 


190  ERECTION  OF  SUSPENSION  BRIDGES 

months  were  lost  in  waiting  for  steel  delivery.     The  bridge  con- 
tains 1600  tons  of  structural  steel  and  250  tons  of  cables. 

The  4<x>-foot-span  suspension  bridge  at  Massena,  New  York, 
(Fig.  30;  H.  D.  Robinson,  Consulting  Engineer)  containing  400 
tons  of  steel,  was  erected  complete  in  six  months. 


APPENDIX 
DESIGN  CHARTS  FOR  SUSPENSION  BRIDGES 

INTRODUCTION. — To  expedite  the  proportioning  or  checking  of 
suspension  bridges,  the  author  has  devised  the  three  charts  which 
are  presented  in  this  Appendix.  These  charts  give  directly  the 
maximum  and  minimum  moments  and  shears  in  the  stiffen- 
ing truss,  throughout  the  main  and  side  spans.  The  charts 
are  constructed  for  the  usual  form  of  construction,  parabolic 
cable  with  two-hinged  stiffening  truss;  and  they  cover  both 
types: 

Type  2F— Free  Side  Spans  (Straight  Backstays). 

Type  2S — Suspended  Side  Spans  (Curved  Backstays). 

To  use  the  charts,  it  is  simply  necessary  to  calculate  N, 
which  is  a  constant  for  any  given  structure.  This  constant  N 
is  denned  by  Eq.  (125)  or  (167),  Chapter  I;  the  formulas  for  N 
are  also  reproduced  on  the  charts.  In  these  formulas: 

/  =  moment  of  inertia  of  the  truss,  main  span; 
1 1  =  moment  of  inertia  of  the  truss,  side  span; 
A  =area  of  cable  section,  main  span* 
A  i  =  area  of  cable  section,  side  span ; 
E  =  coefficient  of  elasticity  for  truss; 
Ec= coefficient  of  elasticity  for  cable; 

/=  cable  sag,  main  span; 
/i  =  cable  sag,  side  span; 

/  =  main  span  of  cable  (c.  to  c.  of  towers) ; 

/'  =  main  span  of  truss  (c.  to  c.  of  bearings); 

/i  =  side  span  of  truss  (c.  to  c.  of  bearings) ; 

/2  =  side  span  of  cable  (tower  to  anchorage) ; 
ai  =  inclination  of  cable  chord  in  side  span. 
191 


192 


APPENDIX 


DESIGN  CHARTS  FOR  SUSPENSION  BRIDGES  193 

The  value  of  N  is  usually  about  1.70  for  the  case  of  free  side 
spans  (Type  2F),  and  about  1.80  for  the  case  of  suspended  side 
spans  (Type  25). 

For  the  case  of  suspended  side  spans  (Type  25)  it  is  also 
necessary  to  figure  the  ratio-product  ii^v,  where- 


~I?       T     -/• 

This  ratio-product  is  also  a  constant  for  any  given  structure. 
(It  is  equal  to  zero  when  the  backstays  are  straight,  Type  2F. 
For  Type  25  we  may  usually  assume  i=i,  and  v  =  r2,  so  that 
ir*v  =  r5,  approximately.) 

With  the  values  of  the  two  constants  N  and  ir*v  known,  the 
maximum  and  minimum  moments  and  shears  for  all  points  in 
main  and  side  spans  may  be  read  directly  from  the  charts,  thus 
dispensing  with  the  usual  laborious  computations. 

Chart  I.  —  Bending  Moments  in  Main  Span.  —  This  chart  gives 
the  governing  bending  moments  throughout  the  main  span.  The 
upper  curves  (for  different  values  of  N)  give  the  maximum  bending 
moments,  and  the  lower  curves  (for  different  values  of  N)  give  the 
minimum  bending  moments.  No  correction  is  required  except  for 
minimum  moments  in  the  case  of  suspended  side  spans  (Type 
25).  The  corrections  for  this  case  are  given  by  the  parabolic 

curves  plotted  below  the   axis   (for   different  values  of  -—  \. 

These  corrections,  like  the  minimum  moments,  are  negative  in 
sign,  and  the  two  should  therefore  be  added  arithmetically. 
(These  corrections  represent  the  effect  of  load  covering  both 
side  spans.) 

The  values  of  Total  M  (for  full  loading  of  all  spans)  may  be 
obtained,  if  desired,  by  arithmetically  subtracting  the  corrected 
Min.  M  from  Max.  M.  Total  M  for  load  covering  the  main 
span  alone  may  be  obtained  by  subtracting  the  uncorrected 
Min.  M  from  Max.  M.  The  resulting  values,  in  either  case, 
would  be  represented  by  parabolas  above  the  axis. 

Chart  II.  —  Shears  in  Main  Span.  —  This  chart  gives  the  gov- 
erning shears  throughout  the  main  span.  The  upper  curves  (for 


194 


APPENDIX 


DESIGN  CHARTS  FOR  SUSPENSION  BRIDGES  195 

different  values  of  N)  give  the  maximum  shears,  and  the  lower 
curves  (for  different  values  of  N)  give  the  minimum  shears.  No 
correction  is  required  except  for  minimum  shears  in  the  case  of 
suspended  side  spans  (Type  25).  The  corrections  for  this 
case  are  given  by  the  straight  lines  plotted  below  the  axis 

(ifsv\ 
for  different  values  of  -js-l-     These  corrections  are  of  the 

same  algebraic  sign  as  the  minimum  shears,  and  the  two  should 
therefore  be  added  arithmetically.  (These  corrections  represent 
the  effect  of  load  covering  both  side  spans.) 

On  this  chart,  the  plus  sign  indicates  a  shear  upward  on  the 
outer  side  and  downward  on  the  inner  side  of  a  section;  the 
minus  sign  indicates  a  shear  in  the  opposite  direction. 

The  values  of  Total  V  (for  full  loading  of  all  spans)  may  be 
obtained,  if  desired,  by  arithmetically  subtracting  the  corrected 
Min.  V  from  Max.  V.  Total  V  for  load  covering  the  main  span 
alone  may  be  obtained  by  subtracting  the  uncorrected  Min.  V 
from  Max.  V.  The  resulting  values,  in  either  case,  would  be 
represented  by  radiating  straight  lines  above  the  axis. 

Chart  III. — Moments  and  Shears  in  Side  Spans. — This  chart 
gives  the  governing  stresses  throughout  a  side  span. 

In  the  left-hand  diagram,  the  upper  parabolic  curves  (for 

different  values  of  — -  )  give  the  maximum  bending  moments. 

(These  curves  represent  the  effect  of  load  covering  the  given 
side  span.)  The  lower  parabolic  curves  (for  different  values  of 

T  ~\~'L/1>ti}\ 

— — — J  give  the  minimum  bending  moments.     (These  curves 

represent  the  effect  of  load  covering  the  two  other  spans.) 
The  values  of  Total  M  (for  load  covering  all  three  spans)  may 
be  obtained,  if  desired,  by  arithmetically  subtracting  Min.  M 
from  Max.  M .  The  resulting  values  would  be  represented  by 
flat  parabolas  above  the  axis. 

In  the  right-hand  diagram,  the  upper  curves  (for  different 

£0FWM\ 

values  of  —  j  give  the  maximum  shears.  (These  curves 
represent  the  effect  of  load  covering  the  given  side-span.)  The 


196 


APPENDIX 


DESIGN  CHARTS  FOR  SUSPENSION   BRIDGES  197 

lower  curves  (for  different  values  of  *     *      j  give  the  minimum 

shears.  (These  curves  represent  the  effect  of  load  covering  the 
two  other  spans.)  The  values  of  Total  V  (for  load  covering  all 
three  spans)  may  be  obtained,  if  desired,  by  arithmetically 
subtracting  Min.  V  from  Max.  V.  The  resulting  values  would 
be  represented  by  radiating  straight  lines  above  the  axis. 

In  this  diagram,  the  plus  sign  indicates  a  shear  upward  on 
the  outer  side  and  downward  on  the  inner  side  of  a  section; 
the  minus  sign  indicates  a  shear  in  the  opposite  direction. 

Chart  III  can  also  be  used  for  a  side  span  not  suspended  from 
the  backstays  (Type  2F),  or  for  any  independent  simple  span. 
The  maximum  bending  moments  produced  by  uniform  load  are 
given  by  the  top  curve  in  the  left-hand  diagram;  the  minimum 
bending  moments  are  zero.  The  maximum  shears  produced  by 
uniform  load  are  given  by  the  top  curve  in  the  right-hand  dia- 
gram ;  the  minimum  shears  are  given  by  the  dotted  continuation 
curve  in  the  same  diagram. 

Where  locomotive  loadings  with  axle-concentrations  are 
specified,  the  equivalent  uniform  loads  are  to  be  used  for  p  in 
these  charts. 


INDEX 

Numbers  refer  to  pages.  Illustrations  are  indicated  by  an  asterisk  (*) 
after  page  number.  For  definitions  of  symbols,  consult  pages  listed  under  the 
index  word  Notation. 


Aare  bridge,  118 

Adjustments,   101-103,   I2I>  l69>   I7°> 

171,  176,  182 

Advantages,  69,  70,  78,  79,  82 
Albert  bridge,  77 
Alloy  steels,  84,  85 
Anchor  chains,  92*,  120-122 

—  girders,  121 

—  plates,  121 

Anchorages,  89,  92*,  94*,  104*,  106*, 
107*,  109*,  118,  119*,  120-124, 
161*,  162 

Anchorage  shafts,  122 

—  stresses,  123, 161*,  162 

—  tunnels,  122 

Anchoring  cables,  91,  92*,  93,  121 

—  strands,  1 74* 

Arrangements    of    cross-sections,    72*, 

73*,  83,  84,  92*,  99*,  in* 

—  of  spans,  72,  74 
Assumptions  for  design,  18,  19 
Attachments,  96,  98,  100 

B 

Backstays,  50*,  51*,  62*,  64*,  72,  79, 

88* 

Balance  beam,  175 
Bearings,  103 

Bending  Moments  (see  Moments) 
Braced  cable  construction,  96,  105 
Braced-chain  bridges,  80-82,  103,  108 

three-hinged  type,  63,  64*,  65* 

two-hinged  type,  65*,  66 

hingeless  type,  67* 


Braced-chain  construction,  64*,  65*, 
67*,  103,  106*,  107*,  109*, 
in* 

Bracing,  77,  78,  91,  no,  112,  113 

Brooklyn  bridge,  70*,  71,  72*,  80,  84,  91, 
101,  168,  170,  171 

Budapest  bridges,  94*,  95* 


Cable  bands,  85,  92*,  96,  98,  99*,  177 

—  connections,  121 

—  curve,  i,  2*,  4,  6*,  7*,  9 

—  deflections,  16,  17 

—  deformations,  n,  12,  14*,  15*,  17 

—  diameter,  90,  149,  150 

—  elongation,  16,  17 

—  estimates,  149 

—  in  side  span,  7* 

—  length,  5,  6,  8,  10,  52 

—  sag,  8 

—  spinning,  172*,  173 

—  squeezing,  177 

—  stresses,  3-5, 8,  9,  11,  13,  126,  134, 

ISIi  I52 

—  tension,  3,  4,  10 

—  unsymmetrical,  6*,  7,  20* 

—  vs.  eyebars,  74-76 

—  weight,  149,  150 

—  wire,  85,  89 

—  wrapping,  90,  149,  150,  169,  183, 

184* 

Cables,  85,  87,  89,  90 
Cannes-Ecluse  bridge,  77* 
Castings,  85 
Catenary,  9,  10 


199 


200 


INDEX 


Center  hinge,  72,  80,  101,  103,  104,* 

105,  113 

Central  loading,  14* 
Chain  construction,  75*,  76,  80,  93,  94*, 

95*,  96,  103,  106*,  107*,  109*, 

in*,  187* 

Charts  for  moments,  192*,  196* 
—  for  shears,  193,  194*,  196* 
Chord  stresses,  128 
Cincinnati  bridge,  71 
Clamping,  87 
Clark's  bridge,  94* 
Classes,  71,  72,  78-81 
Classification,  19,  71,  72 
Clifton  bridge,  187 
Closed  sockets,  98 
Coefficient  of  elasticity,  86,  87 
Cologne  bridge,  53*,  74,  115,  118 

erection,  188 

Common  theory,  19 

Compacting  cables,  177 

Comparison  of  types,  78,  79,  96 

Connections,  96,  98,  100,  121 

Continuous  type  (see  Hingeless  type) 

Cradles,  169 

Cradling  of  cables,  78,  90,  91 

Crescent  type,  82 

Cross-sections,  72*,  73*,  83, 84, 92*,  99*, 

106*,  in*,  134* 

Cumberland  R.  bridge,  170,  184,  185 
Curve  of  cable,  i,  2*,  4,  6*,  7*,  9 


Danube  bridges,  94*,  95* 
Deflections  of  cable,  16,  17 

—  of  truss,  49-51 

Deformations  of  cable,  n,  12,  14*,  15*, 

17 

Delaware  River  bridge,  33* 
Depth  of  truss,  83,  102,  103,  108 
Design  assumptions,  18,  19 

—  charts,  191-197 

—  computations,  125, 134, 144, 149, 162 
Details,  73*,  02*,  94*,  106*,  107*,  109* 

134* 

Detroit- Windsor  bridge,  50*,  118,  123 
Diagonal  stays,  70*,  72*,  76,  77* 


Diameter  of  cable,  90,  149,  150 
Displacement  of  crown,  15* 

—  of  saddle,  17 

E 

Eads'  type,  81 

Economic  proportions,  82,  83,  102 

Elastic  coefficient,  86,  87 

Elizabeth  bridge,  71,  74,  85,  95*, 
115,  118 

erection,  186,  187* 

Elongation  of  cable,  16,  17 

Equalizers,  100 

Equilibrium  polygon,  2*,  20*,  61* 

Erection,  163,  164*,  166*,  168*,  172*, 
174*,  178*,  179*,  180*,  181*, 
182*,  184*,  185*,  186*,  187* 

—  adjustments,  169-171, 182, 183 

—  calculations,  169-171,  183 

—  equipment,  165,  173,  175-177,  i79> 

183 

—  force,  165,  176,  177,  181,  183 

—  of  cables,  172*,  173 

—  of  foot  bridges,  166*,  167 

—  of  towers,  163,  164* 

—  of  trusses,  178*,  179*,  180*,  181*, 

182* 

—  records,  165,  175,  177,  181,  186,  189, 

190 

Estimates,  149,  150 
Exact  theory,  19 
Eyebar  bridges,  74,  75*,  76,  94*,  95*, 

96,  106*,  109*,  in*,  187* 

—  chain  erection,  186, 187*,  188,  189 

—  construction,  74,  75* 


Falsework,  186*,  187* 
Fidler  truss,  81,  108,  109* 
Floor  beams,  84, 92* 
Floor  system,  178* 
Florianopolis  bridge,  134* 
Footbridge  cables,  165,  166*,  167 
—  erection,  166*,  167 
Footbridges,  166*,  168*,  186* 
Forces  acting  on  tower,  146,  147 

on  truss,  20* 

Form  of  cable,  i,  2*,  4,  6*,  7*,  9 


INDEX 


201 


Frankfort  bridge,  80,  81,  104* 

Freiburg  bridge,  73* 

Functions  in  formulas,  39*,  42,  43* 


Galvanizing,  89 
Gisclard  system,  77 
Gotha  foot  bridge,  119* 
Grand  Ave.  bridge,  71 
Graphs  for  formulas,  39*,  43* 
Guide  wires,  169, 170 

H 

H-curves,  23,  28*,  31*,  36*,  46*,  58*, 

152,153 

Hangers,  96,  100 
Hauling  towers,  167,  168* 
Hingeless  type,  53*,  54,  55,  58*,  61*, 

67,  102,  in*,  150 

horizontal  tension,  57-59 

• influence  lines,  58* 

moments,  59,  60,  61 

moments  at  towers,  56 

shears,  55 

1 temperature  stresses,  61 

Hinges,  102,  103,  105,  113 
Horizontal  displacement,  15* 
Horizontal  tension  4,  8,  10,  63,  66,  126, 

i34,  152 

from  temperature,  48 

—  two-hinged  type,  26,    27,  33-35, 

37,38 

hingeless  type,  57-59 

Hudson  River  bridge,  (Frontispiece)*, 

71,74,  76,82,110,  in*,  115 


Influence  lines,   23,   24,   25,   28*,  31*, 

36*,  46*,  58*,  152,  153 
Inspection,  118,  122 

K 

Kingston  bridge,  71,  116,  118 

erection,  165,  171,  181,  185*,  189 

Knuckles,  85,  92*,  117,  118,  122 


Lambeth  bridge,  80,  105 
Lateral  bracing,  77,  91 
Length  of  cable,  5,  6,  8,  10,  52 
Limiting  spans,  76,  83 
Loading,  134* 
Loads,  90 

—  on  tower,  146,  147 
Locked  wire  cables,  88 
London  bridges,  80,  105 

M 

Main  span  stresses,  136,  139,  153,  156, 

192*,  193,  194* 
Maintenance,  118,  122 
Manhattan  bridge,  71,  85,  91,  97*,  99*, 

108,  118,  119* 
erection,  163, 164*,  166*,  167, 171, 

172*,    173,    174*,    177,    178*, 

179*,   180*,   181*,  182*,  184* 
Masonry,  114,  120,  123 
Massena  bridge,  86*,  190 
Materials,  84,  85 
Maximum  moments,  28*,  29,  30,  36* 

—  shears,  31*,  32,  46* 
Moment  charts,  192*,  196* 

—  diagrams,  28*,  36*,  58* 
Moments,  21,  20*,  22,  24,  28*,  36*,  58* 

—  in  stiffening  truss,  127,  129,  136,  139, 

153-156,  158 
Movement  of  saddles,  17,  145,  146 

—  of  towers,  145,  146 
Multiple  spans,  74 

N 

Niagara  suspension  bridge,  101,  103 
North  River  bridge,  (Frontispiece}*  71, 

74,  76,  82,  no,  in*,  115 
Notation,  i,  34,  35,  37,  42,  71,  125,  134, 

150,  151,  191 
-,2*,6*,7*,i3*,2o*,36*,6i* 

O 

Ohio  River  bridge,  71 
Open  sockets,  98 
Ordish  system,  77 


202 


INDEX 


Parabolic  bottom  chord,  80,  106* 

—  cable,  4,  6*,  7* 

—  center  line,  80,  109* 

—  coefficients,  127 

—  top  chord,  80,  104*,  107* 
Parallel  wire  cables.  87,  89,  169 
Patent  cables,  88 
Philadelphia-Camden  bridge,  33* 
Pittsburgh  bridges,  71,  74,  81,  82,  106*, 

108,  no 

Plant,  165,  173,  175,  176,  179 
Point  bridge,  71,  81,  106*,  108 
Prague  bridge,  77 
Proportions,  83,  102,  103,  108 
Protection  of  cables,  90 
Pulleys,  116 


Quebec  designs,  71,  74,  81,  107*,  108, 
109*,  118,  189 


Reaction  girders,  120 

Reactions,  161*,  162 

Rel'ff  of  wind  load,  133,  144 

Resultants,  161*,  162 

Rhine  bridge,  53* 

Roadways,  84,  98,  99* 

Rocker  towers.  50*,  53*,  95*,  107*,  109* 

115,  JI7>  IJ8 
Rockers,  117,  118 
Roebling,  74,  169 
Rollers,  116-118 

Rondout  bridge  (see  Kingston  bridge) 
Rope  strand  cables,  150 
Runners,  181 
Rusting  of  cables,  90 


Saddle  movement,  17,  145,  146 
Saddles,  85,  92*,  94*,  104*,  109*,  115, 

116 

Safety,  70,  118 
Sag  of  cable,  8 
Sag  ratio,  5,  n,  76,83 


Section  sheet,  134* 

Seizing,  174,  175 

Seventh  St.  bridge,  71,  74,  82,  no 

Shafts,  122 

Shear  charts,  194*,  196* 

—  diagrams,  31*,  46* 
Shears,  3,  20*,  22,  25 

—  in  stiffening  truss,  130,  131,  139-142, 

156-158,  160 
Sheave  towers,  168* 
Siamese  Railways,  bridge  for,  75* 
Side  span  cable,  7* 

—  stresses,    138,    141,    158,    160,    195, 

196* 

Side  spans,  72,  79 
Sizes  of  wire,  89 
Sliding,  162 
Sockets  85,  92*,  93,  96,  98,  99*,  177, 

178 
Span  arrangements,  72,  74 

—  limits,  76,  83 

Spandrel  braced  types,  So,  81, 104*,  107* 

Splicing  wires,  90,  92*,  99* 

Spinning  cables,  172*,  173 

Squeezing  cables,  177 

St.  Louis  bridge,  71 

Stability,  162 

Steel  towers,  114 

Stiffened  suspension  bridges,  18 

Stiffening,  72*,  76,  77*,  78 

—  trusses,  101 

—  stresses,  20* 

Straight  backstays,  50*,  51*,  62*,  64*, 
88*,  125 

—  bottom  chord,  80,  104* 
Strand  bridle,  1 74* 

—  legs,  170 

-  shoes,  85,  92*,  93,  99*,  172*,  174 

Strands,  89,  90,  93 

Strength,  84-86,  90 

Stress  sheet,  134* 

Stresses  in  anchorage,  123,  161*,  162 

-in  cables,  3-5,  8,  9,  11,  13,  126,  134, 

135,  151,  152 

—  in  chords,  1 28 

—  in  towers,  14,  147,  148 

—  in  truss,  127,  130,  136,  138,  139,  141, 

153,  156,  158, 160 


INDEX 


203 


Suspender  connections,  96,  100 

—  erection,  177,  178* 

—  forces,  20*,  21,  27 
Suspenders,  85,  92*,  09*,  100,  177, 
Suspension  details,  73*,  99* 
Sway  bracing,  no,  112,  113 


Table  I  (functions),  42 

Temperature  stresses,  48,  61,  126,  129, 

132,  142,  143 
Tension  in  cable,  3,  4,  10 
Three-hinged  type,   26,   27,   28*,  64*, 

65*,  106* 

influence  lines,  28*,  31 

moments,  28*,  29,  30 

shears,  30,  31*,  32 

Tiber  bridge,  108 

Tilting,  162 

Time  required,  erection,  181,  186,  189, 

190 

Tower  bridge,  108 
Tower  calculations,  144 

—  erection,  163,  164* 

—  loads,  146,  147 

—  movement,  145,  146 

—  stresses,  14,  147,  148 

Towers,  85,  86*,  99*,  104*,  106*,  109*, 

113-115,  134*,  144,  164* 
Traveling  rope,  172*,  173 

—  sheaves,  172*,  174 

—  wheels,  172*,  174 

Truss  depth,  83,  102,  103,  108 

—  erection,  178*,  179*,  180*,  181*,  182* 
Tunnels,  122 

Two-hinged  type,  33*,  36*,  46*,  50*, 

51*,  65*,  134* 

deflections,  49-51 

horizontal  tension,  33-38,  40,  51, 

52 

influence  lines,  36*,  46* 

moments,  36*,  41,  44,  45,  52 

shears,  45,  46*,  47,  52 

temperature  stresses,  48 

Type  05,  67*,  80 

—  05P,  82,  1 10,  in* 

—  OF,  62*,  102 


Type  OFE,  95* 

-  05,  53*,  58*,  61*,  78,  102,  150 

-  15,  79 

—  25,  65*,  80 

—  2BF,  82 

—  2BH,  80,  81 

—  25P,  82,  1  10 

—  255,  65* 

—  25F,  81,  112 

o7*,  108 


-  2F,  50*,  51*,  75*,  79,  102,  125,  191, 

197 

-2FZ?,77* 
-2FE,75* 
-25,  33*,  36*,  46*,  79,  86*,  97*,  99*, 

102,   134*,   191 

—  35,  64*,  65*,  80 

—  35C5,  iog*,  1  10,  112 

—  35F,  64* 

—  3BH,  103,  104,  112 

—  35L,  81 

—  3BLF,  106*,  108^110 

—  355,  65* 

—  3BUH,  80,  81 

—  3F,  28*,  31*,  79,  ioi 

—  357?,  70*,  72* 

Types  of  suspension  bridges,  18,  19,  71, 
72,  78-81 

U 

Unit  stresses,  134* 

Unstiffened  suspension  bridges,  12,  13*, 

14*,  15*,  73*,  94* 
Unsymmetrical  cable,  6*,  7*,  20* 

—  loading,  15* 
Uplift,  120,  123 

V 

Vertical  deflection,  14* 
Vierendeel  girder,  ioi,  in*,  112 
Villefranche  bridge,  77 

W 

Web  systems,  ioi 

Width  of  bridge,  83 

Williamsburg  bridge,  71,  84,  88*,  91, 

92* 
--  erection,  165,  168,  171,  175,  184, 

186* 


204 


Wind  bracing,  no,  112,  113 

—  cables,  74,  106*,  113 

—  chords,  78,  112 

—  loads,  133 

—  stresses,  132,  133,  144,  148 
Wire  for  cables,  89,  90 

Wire  rope  bridges,  77*,  86* 

—  cables,  91,  184,  185 

—  link  bridges,  107* 


INDEX 


Wire  ropes,  85-87,  93 

—  splice,  90,  92*,  99* 
—  wrapping,  90 

Wrapping,  90,  149,  150 

—  machine,  183,  184* 


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